IN   MEMORIAM 
FLOR1AN  CAJORI 


>mBb 

■'•■•  ■■    -    KJ:: 
■S.Vi:_.a3-- 

*°  A 


%;<sh 


^sr^rxy 


AN 


ELEMENTARY   TREATISE 


PLANE  &  SPHERICAL  TRIGONOMETRY, 

WITH    THEIR    APPLICATIONS    TO 

NAVIGATION,  SURVEYING,  HEIGHTS  &  DISTANCES, 
AND    SPHERICAL   ASTRONOMY, 

AND    PARTICULARLY    ADAPTED    TO    EXPLAINING    THE 

CONSTRUCTION    OF   BOWDITCIPS   NAVIGATOR,    AND 
THE  NAUTICAL   ALMANAC. 


By  BENJAMIN   PEIRCE,  A.  M., 

University  Professor  of  Mathematics  and  Natural  Philosophy  in  Harvard  University. 


BOSTON: 
JAMES    MUNROE    AND    COMPANY. 

M  DCCC  XL. 


Entered  according  to  act  of  Congress,  in  the  year  1840,  by  James  Munroe  and  Com- 
pany, in  the  Clerk^s  Office  of  the  District  Court  of  the  District  of  Massachusetts, 


CAMBRIDGE  PRESS: 
METCALF,  TORRY,  AND  BALLOU. 


*V 


7>3? 


CONTENTS. 


PLANE    TRIGONOMETRY. 

Chap.  Page- 

I.    General  Principles  of  Plane  Trigonometry    ...  3 

II.     Sines,  Tangents,  and  Secants     .....  6 

III.  Right  Triangles .20 

IV.  General  Formulas 27 

V.     Values  of  the  Sines,  Cosines,  Tangents,  Cotangents, 

Secants,  and  Cosecants  of  certain  Angles    ...  37 

VI.    Oblique  Triangles 47 

NAVIGATION  AND  SURVEYING. 

I.    Plane  Sailing 67 

II.     Traverse  Sailing 76 

III.  Parallel  Sailing 80 

IV.  Middle  Latitude  Sailing 83 

V.    Mercator's  Sailing           ...*...  90 

VI.    Surveying 103 

VII.    Heights  and  Distances 115 

SPHERICAL  TRIGONOMETRY. 

I.    Definitions .  129 

II.    Right  Triangles 134 

III.    Oblique  Triangles 156 


o?is>€a»3*»ii 


IV  CONTENTS. 

SPHERICAL  ASTRONOMY. 

CHAP.  Page. 

I.  The  Celestial  Sphere  and  its  Circles  ....  191 

II.  TJhe  Diurnal  Motion 196 

III.  The  Meridian 210 

IV.  Latitude 223 

V.  The  Ecliptic  .        .        .        .        .        .        .        .261 

VI.  Precession  and  Nutation 275 

VII.  Time 293 

VIII.  Longitude 309 

IX.  Aberration 340 

X.  Refraction 356 

XI.  Parallax 367 

XII.  Eclipses 383 


ERRATA. 

\  11,  ] 

13, 

•  -2,  for 
2, 

6.7914  read  6.7935 
BCT            ACT 

P. 

307, 

I.  21, 

for 

9h17m  Qs  r>    ^427n52 

15 

13, 

0.00354 

0.00364 

325, 

—3, 

fiar.43 

fig.  40. 

56, 

13, 

AB 

AC 

338, 

4, 

51°  2 

51^43 '39 

186, 

-5, 

tangent 

sine 

8&  £ 

, 

dele 

R.  A. 

206, 

15, 

fig.  2 

fig.  35 

356, 

-5, 

for 

dec.    read 

upper 

an, 

7, 

fig.  37 

fig.  38 

369, 

— 1, 

attain 

obtain 

235, 

15, 

ZSD 

ZSP 

370, 

6, 

AE 

AL 

264, 

8, 

long.  90° 

long.  —90- 

14, 

ABR 

BAR 

277, 

15  &  16, 

PZ 

h& 

-3, 

RAB 

RB 

278, 

10, 

APAX 

*I 

384, 

7, 

OPO' 

opo1 

—  1, 

NAAX 

SPS' 
4A36w  Qs 

390, 

3, 

JVg 

Mg 

315, 
307, 

-6, 
17, 

SS'D 

7/*23w51s 

401, 

4, 
15, 

MUG 
61 

JVMg 
63 

PLANE    TRIGONOMETRY. 


Digitized  by  the  Internet  Archive 

in  2008  with  funding  from 

Microsoft  Corporation 


http://www.archive.org/details/elementarytreatiOOpeirrich 


PLANE    TRIGONOMETRY. 


CHAPTER   I. 


GENERAL    PRINCIPLES    OP    PLANE    TRIGONOMETRY. 

1.  Trigonometry  is  the  science  which  treats  of 
angles  and  triangles. 

2.  Plane  Trigonometry  treats  of  plane  triangles. 
[B.  p.  36.]* 

3.  To  solve  a  Triangle  is  to  calculate  certain  of 
its  sides  and  angles  when  the  others  are  known. 

It  has  been  proved  in  Geometry  that,  when  three  of  the  six 
parts  of  a  triangle  are  given,  the  triangle  can  be  constructed, 
provided  one  at  least  of  the  given  parts  is  a  side.  In  these 
cases,  then,  the  unknown  parts  of  the  triangle  can  be  deter- 
mined geometrically,  and  it  may  readily  be  inferred  that  they 
can  also  be  determined  algebraically. 

But  a  great  difficulty  is  met  with  on  the  very  threshold  of 
the  attempt  to  apply  the  calculus  to  triangles.  It  arises  from 
the  circumstance,  that  two  kinds  of  quantities  are  to  be  intro- 

*  References  between  brackets,  preceded  by  the  letter  B.,  refer  to 
the  pages  in  the  stereotype  edition  of  Bowditch's  Navigator. 


4  PLANE  TRIGONOMETRY.  [CH.  I. 

Solution  of  all  triangles  reduced  to  that  of  right  triangles. 

duced  into  the  same  formulas,  sides,  and  angles.  These  quan- 
tities are  not  only  of  an  entirely  different  species,  but  the  law  of 
their  relative  increase  and  decrease  is  so  complicated,  that 
they  cannot  be  determined  from  each  other  by  any  of  the  com- 
mon operations  of  Algebra. 

4.  To  diminish  the  difficulty  of  solving  triangles  as 
much  as  possible,  every  method  has  been  taken  to  com- 
pare triangles  with  each  other,  and  the  solution  of  all 
triangles  has  been  reduced  to  that  of  a  Limited  Series 
of  Right  Triangles. 

a.  It  is  a  well  known  proposition  of  Geometry,  that,  in  all 
triangles,  which  are  equiangular  with  respect  to  each  other, 
the  ratios  of  the  homologous  sides  are  also  equal.  [B.  p.  12.] 
If,  then,  a  series  of  dissimilar  triangles  were  constructed  con- 
taining every  possible  variety  of  angles :  and,  if  the  angles 
and  the  ratios  of  the  sides  were  all  known,  we  should  find  it 
easy  to  calculate  every  case  of  triangles.  Suppose,  for  instance, 
that  in  the  triangle  ABC  (fig.  1.),  the  sides  of  which  we  shall 
denote  by  the  small  letters  a,  b,  c,  respectively  opposite  to  the 
angles  A,  Bf  C,  there  are  given  the  two  sides  b  and  c 
and  the  included  angle  A,  to  find  the  side  a  and  the  angles 
B  and  C.  We  are  to  look  through  the  series  of  calcu- 
lated triangles,  till  we  find  one  which  has  an  angle  equal 
to  A,  and  the  ratio  of  the^  including  sides  equal  to  that  of 
b  and  c.  As  this  triangle  is  similar  to  ABC,  its  angles  and 
the  ratio  of  its  sides  must  also  be  those  of  the  triangle  ABC, 
which  is  therefore  completely  determined.  For,  to  find  the 
side  a,  we  have  only  to  multiply  the  ratio  which  we  have  found 

of  b  to  a,  that  is,  the  fraction  -  by  the  side  b  or  the  ratio  -  by 
the  side  c. 


§  4.]  GENERAL  PRINCIPLES. 


Solution  of  all  triangles  reduced  to  that  of  right  triangles. 

6.  A  series  of  calculated  triangles  is  not,  however,  needed 
for  any  other  than  right  triangles.  For  every  oblique  triangle 
is  either  the  sum  or  the  difference  of  two  right  triangles ;  and 
the  sides  and  angles  of  the  oblique  triangle  are  the  same  with 
those  of  the  right  triangles,  or  may  be  obtained  from  them  by 
addition  or  by  subtraction.  Thus  the  triangle  ABC  is  the 
sum  (fig.  2.)  or  the  difference  (fig.  3.)  of  the  two  right  trian- 
gles ABP  and  BpC.  In  both  figures  the  sides  AB,  BC9 
and  the  angle  A  belong  at  once  to  the  oblique  and  the  right 
triangles,  and  so  does  the  angle  BCA  (fig.  2.)  or  its  supple- 
ment (fig.  3.) ;  while  the  angle  ABC  is  the  sum  (fig.  2.),  or, 
the  difference  (fig.  3.)  of  ABP  and  PBC;  and  the  side  AC 
is  the  sum  (fig.  2.),  or  the  difference  (fig.  3.)  of  AP  and  PC. 

c.  But,  as  even  a  series  of  right  triangles,  whiqh  should  con- 
tain every  variety  of  angle,  would  be  unlimited,  it  could  never 
be  constructed  or  calculated.  Fortunately,  such  a  series  is 
not  required ;  and  it  is  sufficient  for  all  practical  purposes  to 
calculate  a  series  in  which  the  successive  angles  differ  only  by 
a  minute,  or,  at  the  least,  b  ya  second.  The  other  triangles  can 
be  obtained,  when  needed,  by  that  simple  principle  of  inter- 
polation made  use  of  to  obtain  the  intermediate  logarithms 
from  those  given  in  the  tables. 


1* 


PLANE  TRIGONOMETRY.  [cH.   II. 


Sine,  tangent,  secant. 


CHAPTER    II. 

SINES,  TANGENTS,  AND  SECANTS. 

5.  Confining  ourselves,  for  the  present,  to  right  tri- 
angles, we  now  proceed  to  introduce  some  terms,  for 
the  purpose  of  giving  simplicity  and  brevity  to  our 
language. 

The  Sine  of  an  angle  is  the  quotient  obtained  by 
dividing  the  leg  opposite  it  in  a  right  triangle  by  the 
hypothenuse. 

Thus,  if  we  denote  (fig.  4.)  the  legs  BC  and  AC  by  the 
letters  a  and  6,  and  the  hypothenuse  AB  by  the  letter  A,  we 
have. 

sin.  A  =  -,     sin.  B  =  T.  (1) 

h  li 

6.  The  Tangent  of  an  angle  is  the  quotient  obtained 
by  dividing  the  leg  opposite  it  in  a  right  triangle,  by 
the  adjacent  leg. 

Thus,  (fig.  4.), 

tang.  A  =  -,     tang.  B=  -.  (2) 

7.  The  Secant  of  an  angle  is  the  quotient  obtained 
by  dividing  the  hypothenuse  by  the  leg  adjacent  to  the 
angle. 


§  10.]  SINES,  TANGENTS,  AND  SECANTS. 


Cosine,  cotangent,  cosecant. 


Thus,  (fig.  4.) 


sec.  A  =  T,     sec.  B  =  -. 
b  a 


(3) 


8.  The  Cosine,  Cotangent,  and  Cosecant  of  an  angle 
are  respectively  the  sine,  tangent,  and  secant  of  its 
complement. 


9.  Corollary.  Since  the  two  acute  angles  of  a  right 
triangle  are  complements  of  each  other,  the  sine,  tan- 
gent, and  secant  of  the  one  must  be  the  cosine,  cotan- 
gent, and  cosecant  of  the  other. 

Thus,  (fig.  4.), 


sin. 


cos. 


A  =  cos. 


sin. 


tang.   A  =  cotan.I2  == 


a 


cotan.  A  =  tang.  B  =  - 


A  —  cosec.2?  = 


cosec.  A  =  sec.     B  = 


(4) 


10.  Corollary.  By  inspecting  the  preceding  equations 
(4),  we  perceive  that  the  sine  and  cosecant  of  an  angle 
are  reciprocals  of  each  other ;  as  are  also  the  cosine 
and  secant,  and  also  the  tangent  and  cotangent. 


PLANE  TRIGONOMETRY. 


[CH.  II. 


To  find  the  tangent. 


So  that 


cosec.  A  X  sin.     A  =  -  X  t  —  — 7  =  1 
a       A        a  A 


sec.      .4  X  cos.    ^4 
tang.    A  X  cotan.  -4 


A       b_bh_ 
b*  h-~  bh~ 

a        b  ab 

0       a       ab 


whence 


cosec.  A  =22  - 7,  or  sin.     A  == T 

sin.  .4  cosec.  A 


sec. 


1  A  1 

cotan.  A  — -,  or  tang.  A  = 


tan.  ^4 


cotan. A 


(5) 


: --,  or  cos.    A  i= j-  ^        (6) 

cos.  ^4  sec.  A    {        v  ' 


As  soon,  then,  as  the  sine,  cosine,  and  tangent  of  an  angle 
are  known,  their  reciprocals  the  cosecant,  secant,  and  cotan- 
gent may  easily  be  obtained. 


11.  Problem.     To  find  the  tangent  when   the  sine 
and  cosine  of  an  angle  are  known. 

Solution.   The  quotient  of  sin.  A  divided  by  cos.  A  is,  by 
equations  (4), 

sin.  A       a       b       ah       a 
cos.  A  ~~  h    '   k       bh~~  b' 


But  by  (4) 


hence 


tang.  4  =  -; 


tang.  A  = 


sin.  A 
cos.  A* 


(?) 


§  4.]         SINES,  TANGENTS,  AND  SECANTS.  9 

Sum  of  squares  of  sine  and  cosine. 

12.  Corollary.  Since  the  cotangent  is  the  reciprocal 
of  the  tangent,  we  have 

cotan.  A  =.  — — .  (8) 

sin.  A  v  ' 

13.  Problem.    To  find  the  cosine  of  an  angle  when 
its  sine  is  known. 

Solution.  We  have,  by  the  Pythagorean  proposition,  in  the 
right  triangle  ABC  (fig.  4.) 

a2  +  b2  =z  h2. 

But  by  (4) 

/  •     -V-    ,    /         A-        «2    ,    °2        a2  +b2        h2       • 

or  (sin.  4)2  +  (cos.  4)2  =  1  •  (9) 

that  is,  the  sum  of  the  squares  of  the  sine  and  cosine  is 
equal  to  unity. 

Hence  (cos.  A)2  =  1  —  (sin.  A)2, 


cos.  A  =  VI  —  (sin.  A)2.  (10) 

I 

14.  Corollary.     Since 

h2  —  a2z=  b2, 

we  have  by  (4) 

h2       a2        h2  —  a2        b2      t 
{sec.A)2-(t^g.A)2  =  -^--  =  —^—  =  -  =  h 

or  (sec.  A)2  —  (tang,  ^l)2  =  1 ;  (11) 

whence  (sec.  A)2  ==  1  +  (tang.  A)2. 


10  PLANE  TRIGONOMETRY.  [CH.  II. 

Calculation  of  cosine,  &c. 

15.  Corollary.    Since 

h2  —  b2=za2 

we  have  by  (4) 

axo      /     *        A       h2       b2       h2—l>2       a2       , 

(cosec-4)2  —  (cotan.^4)2  =  — -  = —  =  — _  1, 

v  '        v  '        a2       a2  a2  a2  —    ' 

or  (cosec.  A)2  —  (cotan.  A)2  =  I ;  (12) 

whence  (cosec.  A)2  =  1  -f-  (cotan.  A)2. 

16.  Scholium.  The  whole  difficulty  of  calculating 
the  trigonometric  tables  of  sines  and  cosines,  tangents 
and  cotangents,  secants  and  cosecants  is,  by  the  pre- 
ceding propositions,  reduced  to  that  of  calculating  the 
sines  alone. 

17.     Examples. 

I.  Given  the  sine  of  the  angle  A,  equal  to  0.4568,  calculate 
its  cosine,  tangent,  cotangent,  secant,  and  cosecant. 

Solution.    By  equation  (10) 


cos.  A  =  s/l  —  (sin.  A )2  =  V(l  +  sin..4)(l  — sin.^l). 
1  -+-  sin.  A  =  1.4568  0.16340 

1  —  sin.  A  =  0.5432  9.73496 


(cos.  A)2  2|9.89836 

cos.  A  =  0.8896  9.94918. 

By  (7)  and  (8) 

-        sin.  A  .       cos.  A 

tang.  A  = cotan.  A  =  -: -r. 

cos.  A  sin.  A 


$  17.]  SINES,  TANGENTS,   AND  SECANTS.  11 


Calculation  of  cosine,  &c. 


sin.  A  =  0.4568                    9.65973     (ar.  co.)  10.34027 

cos.  A  =  0.8896  (ar.  co.)  10.05082  9.94918 

tang.  A  =  0.5135                    9.71055     (ar.  co.)  10.28945 

cotan.  A  =  1.9474. 


By  (6) 


sec.  A  = r,  cosec.  A 


cos.  A  sm.  A 

log.  sec.  A  =  —  log.  cos.  A  =  0.05082, 
sec.  A  =  1.1241. 
log.  cosec.  A  =  —  log.  sin.  ,4  =  0.34027, 
cosec.  it  =  2.1891. 

2.  Given  sin.  4.  =0.1111 ;  find  the  cosine,  tangent,  cotan- 
gent, secant,  and  cosecant  of  A, 

Ans.       cos.  A  =  0.9938 

tang.  .4=0.1118 

cotan.  A  =  8.9452 

sec.  A  =  1.0062 

cosec.  4  f=  9.0010. 

3.  Given  sin.  J.  =  0.9891 ;  find  the  cosine,  tangent,  cotan- 
gent, secant,  and  cosecant  of  A, 

A?is.       cos.  A  =  0.1472 

tang.  A  =  6.7173 

cotan.  4  =  0.1489 

sec.  A  =  6.7914 

cosec.  A=  1.0110. 


12  PLANE  TRIGONOMETRY.  [CH.  II. 


Sine,  &c.  in  the  circle  whose  radius  is  unity. 


18.  Theorem.  The  sine  of  an  angle  is  equal  to  the 
perpendicular  let  fall  from  one  extremity  of  the  arc, 
which  measures  it  in  the  circle,  whose  radius  is  unity, 
upon  the  radius  passing  through  the  other  extremity. 

Proof.  Let  BCA  (fig.  5.)  be  the  angle,  and  let  the  radius 
of  the  circle  AB  A1  A  be 

AC  =  unity  =  1. 

Let  fall,  on  the  radius  AC,  the  perpendicular  BP,  and  we 
have  by  §  5,  in  the  right  triangle  BCP, 

sin.BCP  =  |J  =  ^  =  BF. 

19.  Theorem.  In  the  circle  of  which  the  radius  is 
unity,  the  cosine  of  an  angle  is  equal  to  the  portion  of 
the  radius,  which  is  drawn  perpendicular  to  the  sine, 
included  between  the  sine  and  the  centre. 

Proof.  For  if  BCA  (fig.  5.)  is  the  angle,  we  have,  by  §  9, 

C P         CT> 
cos.  BCA  =  ~  =  ZZL  =  CP. 

20.  Theorem.  In  the  circle  of  which  the  radius  is 
unity,  the  secant  is  equal  to  the  length  of  the  radius 
drawn  through  one  extremity  of  the  arc  which  measures 
the  angle,  and  produced  till  it  meets  the  tangent  drawn 
through  the  other  extremity. 

The  trigonometric  tangent  is  equal  to  that  portion 
of  the  tangent,  drawn  through  one  extremity  of  the  arc, 
which  is  intercepted  between  the  two  radii  which  termi- 
nate the  arc. 


§  22.]        SINES,  TANGENTS,  AND  SECANTS.  13 

Sine,  &c.  in  this  and  other  systems. 

Proof.  If  CB  (fig.  5.)  is  produced  to  meet  the  tangent  A 
at  T,  we  have,  by  (2)  and  (3),  in  the  right  triangle  BCT, 

sec.  BCA  =  ~  -  -^-  =Cr 

tang.  BCA  =  -^  =  —  =  ^T. 

21.  Scholium.  The  preceding  theorems  (18-20), 
have  been  adopted  by  most  writers  upon  trigonometry, 
as  the  definitions  of  sine,  cosine,  tangent,  and  secant, 
except  that  the  radius  of  the  circle  has  not  been  limited 
to  unity.    [B.  p.  6.] 

By  not  limiting  the  radius  to  unity  the  sines,  &c.  have  not 
been  fixed  values,  but  have  varied  with  the  length  of  the 
radius ;  whereas  their  values,  in  the  system  here  adopted,  are 
the  fixed  ratios  of  their  values  as  ordinarily  given  to  the  radius 
of  the  circle  in  which  they  are  measured.  Thus,  if  R  is  the 
radius,  we  have 

sin.,  cos.,  &c.  in  the  common  system  ==  R  X  sin.,  cos.,  &c. 
in  this  system. 


22.  Corollary.  If  the  angle  is  very  small,  as  C  (fig.  6.),  the 
arc  AB  will  be  sensibly  a  straight  line,  perpendicular  to  the 
two  radii  CA  and  CB,  drawn  to  its  extremities,  and  will  sen- 
sibly coincide  with  the  sine  and  tangent ;  while  the  cosine  will 
sensibly  coincide  with  the  radius  CA9  and  the  secant  with  the 
radius  CB. 

Hence,  the  sine  and  tangent  of  a  very  small  angle 
are  nearly  equal  to  the  arc  ivhich  measures  the  angle, 

2 


14  PLANE  TRIGONOMETRY.  [CH.   II. 

Sine,  &c.  of  very  small  angles. 

in  the  circle  the  radius  of  which  is  unity  ;  and  its  cosine 
and  secant  are  nearly  equal  to  unity. 

23.  Problem.   To  find  the  sine  of  a  very  small  angle. 

Solution.  Let  the  angle  C  (fig.  6.)  be  the  given  angle,  and 
suppose  it  to  be  exactly  one  minute.  The  arc  AB  must  in 
this  case  be  ^jsirjir  of  the  semicircumference,  of  which  unity 
or  CA  is  radius.  But  the  value  of  the  semicircumference,  of 
which  unity  is  radius,  has  been  found  in  Geometry  to  be 
3.1415926.     Therefore,  by  §22, 

™,.r=^  =  Hi^  =  o.o<«9.  m 

In  the  same  way  we  might  find  the  sine  of  any  other  small 
angle,  or  we  might,  in  preference,  find  it  by  the  following 
proposition. 

24.  Theorem.  The  sines  of  very  small  angles  are 
proportional  to  the  angles  themselves. 

Proof.  Let  there  be  the  two  small  angles,  BCA  and  B'CA 
(fig.  7.)  Draw  the  arc  ABB1  with  the  centre  C,  and  the  radius 
unity.  Then,  as  angles  are  proportional  to  the  arcs  which 
measure  them, 

BCA :  BCA  =  BA  :  B'A. 

But,  by  §  22, 

sin.  BCA  =f  BA,  sin.  BCA  =  B'  A  ; 
whence 

BCA  :  BCA  =  sin.  BCA  :  sin.  BCA. 

a.  This  proposition  is  limited  to   angles  so  small,  that  their 
arcs  may  be  considered  as  straight  lines.     It  is  found  in  prac- 


§  26.]  SINES,  TANGENTS,  AND  SECANTS.  15 

Sines  of  small  angles. 

tice,  that  the  angles  may  be  as  large  as  two  degrees,  provided 
the  approximations  are  not  carried  beyond  five  places  of  deci- 
mals. The  investigation  of  the  sines  of  larger  angles  requires 
the  introduction  of  some  new  formulas. 


25.    Examples. 

1.  Find  the  sine  of  12'  13",  knowing  that 

sin.  1'  =  0.00029. 

Solution.  By  (28) 

1':  12'  13":  :  sin.  V  :  sin.  12'  13", 

or 

60" :  733"  :  :  0.00029  :  sin.  12'  13". 

Hence 

sin.  12'  13"  =  **  X  0-00029  =  Qm.    Ans 
bO 

2.  Find  the  sine  of  7'  15",  knowing  that 

sin.  V  =  0.00029. 

Ans.     sin.  7'  15"  =  0.00210. 

3.  Find  the  sine  of  2'  31",  knowing  that 

sin.  1'  ==  0.00029. 

Ans.     sin.  2'  31"  =  0.00073. 

26.  Problem.  Given  the  sine  of  any  angle,  to  find 
the  sine  of  another  angle  which  exceeds  it  by  a  very 
small  quantity. 


16  PLANE  TRIGONOMETRY.  [CH.  II. 

Sine  of  an  angle  differing  very  little  from  a  given  angle. 

Solution.  Let  the  given  angle  be  BCA  (fig.  8),  which  we 
will  denote  by  the  letter  31;  and  let  the  angle  whose  sine  is 
required  be  B'CA,  exceeding  the  former  by  the  small  angle 
B'CB,  which  we  will  denote  by  the  letter  m  ;  so  that 

M=  BCA,  m  =B'CB, 

3f+m  =  B'CA. 

From  the  vertex  C  as  a^  centre,  with  the  radius  unity,  de- 
scribe the  arc  ABB'.  From  the  points  B  and  B'  let  fall  BP 
and  B'P'  perpendicular  to  AC. 

We  have,  by  §  18  and  19, 

Sine.  M=BP 
sin.  BCA  =  sin.  (31  -f  m)  =  B'  P' 
cos.  31=  PC; 

Draw  BR  perpendicular  to  B'  P',  and 
B>  P<  =BP  +  BR, 

or 

sin.  (31+  in)  =  sin.  31  +  BR. 

The  triangles  BCP  and  BBR}  having  their  sides  perpen- 
dicular each  to  each,  are  similar,  and  give  the  proportion 

BC:  BB=CP  :  BR. 

But,  by  §  22, 

BB'  ===  sin.  m. 
Hence 

1  :  sin.  m  =  cos.  31 :  BR  ; 
and  BR  =  sin.  m.  cos.  3T, 

which  gives,  by  substitution, 


§  28.]  SINES,  TANGENTS,  AND  SECANTS.  17 

Cosine  of  an  angle  differing  very  little  from  a  given  angle, 
sin.  (M  +  m)  =  sin.  M  +  sin.  m.  cos.  M.  (13) 

27.  Corollary.  If  m  were  1',  (13)  would  become 
sin.  (M  -f-  l'J  =  sin.  iff  -f-  sin.  1'.  cos.  M, 

=  sin.  Jf  +  0.00029  cos.  M.  (14) 

We  may,  by  this  formula,  find  the  sine  of  2'  from  that  of  1', 
thence  that  of  3',  then  of  4',  of  5',  &,c,  to  the  sine  of  angle 
of  any  number  of  degrees  and  minutes. 

28.  Corollary.  We  can,  in  a  similar  way,  deduce  the  value 
of  cos.  (M+  m). 

For,  by  §  19, 

cos.  (if  +  to)  =  P'C=  PC—  PP', 

=  cos.  M—BR. 

But  the  similar  triangles 

BB'R  and  BCP  give  the  proportion 

BC:BB'  =  BP:  BR, 


or 


Hence 


1  :  sin.  m  sss  sin.  Jf :  .RR. 
BR  =z  sin.  m.  sin.  M, 


whence 

cos.  (Jf  -f-  m)  =  cos.  ifcT —  sin.  m.  sin.  i!f,  (15) 

and,  if  we  make  m  =  1',  this  equation  becomes 
cos.  (1/  +  I')  =  cos«  M —  sin.  T#  sin  M, 

ss  cos.  if—  0.00029  sin.  ilif.  (16) 

2* 


18  PLANE  TRIGONOMETRY.  [cH.  If. 


Sine  and  cosine  of  angles. 


29.    Examples. 

1.  Given  the  sine  of  23°  28'  equal  to  0.39822,  to  find  the 
sine  of  23°  29'. 

Solution.  We  find  the  cosine  of  23°  28'  by  (10)  to  be 

cos.  23°  28'  =  0.91729. 
Hence,  by  (14),  making  M  =  23°  28' 

sin.  23°  29'  =  sin.  23°  28'  +  0.00029  cos.  23°  28', 

=  0.39822  +  0.00026, 

=  0.39848. 

Ans.     sin.  23°  29'  —  0.39S48. 

2.  Given  the  sine  and  cosine  of  46°  58'  as  follows, 

sin.  46°  58'  ==  0.73096,  cos.  46°  58'  =  0.68242, 
find  the  sine  of  46°  59'. 

Ans.     sin.  46°  59'  =  0.73116. 

3.  Given  the  sine  and  cosine  of  11°  10'  as  follows, 

sin.  11°  10'  m  0.19366,  cos.  11°  10'  fc=  0.98107, 
find  the  cosine  of  11°  ll7. 

Ans.     cos.  11°  11  =  0.98101. 

30.  By  the  formulas  here  given  a  complete  table  of 
sines  and  cosines  might  be  calculated.  Such  tables 
have  been  actually  calculated  ;  and  table  XXIV.  of  the 
Navigator  is  such  a  table ;  their  logarithms  are  given 
in  table  XXVII.  of  the  Navigator. 


§  30.]  SINES,  TANGENTS,  AND  SECANTS.  19 

Natural  and  artificial  sines.     Radius  of  table. 

The  sines,  cosines,  &c.  of  table  XXIV.  are  called  natural, 
to  distinguish  them  from  their  logarithms,  which  are  some- 
times called  their  artificial  sines,  cosines,  &c. 

The  radius  of  table  XXIV.  is 

10  5  ==  100000, 

so  that  this  table  is,  by  §  21,  reduced  to  the  present  system 
by  dividing  each  number  by  100000,  that  is,  by  prefixing  the 
decimal  point  to  each  of  the  numbers  of  the  table. 

The  radius  of  table  XXVII.  is 

1010z=z  10000000000, 

so  that  this  table  is  reduced  to  the  present  system  by  subtract- 
ing from  each  number  the  logarithm  of  this  radius,  which 
is  10,  that  is  by  subtracting  10  from  each  characteristic. 

The  method  of  using  these  two  tables  is  fully  explained  in 
pp.  33  -  35,  and  p.  390,  of  the  Navigator. 


20  PLANE  TRIGONOMETRY.  [CH.  III. 

Hypothenuse  and  an  angle  given. 


CHAPTER   III. 


RIGHT    TRIANGLES. 


31.  Problem.   To  solve  a  right  triangle^  when  the 
hypothenuse  and  one  of  the  angles  are  known.  [B.  p.  38.] 

Solution.  Given  (fig.  4)  the  hypothenuse  h  and   the  angle 
A,  to  solve  the  triangle. 

First.  To  find  the  other  acute  angle  B,  subtract  the  given 
angle  from  90°. 

Secondly.  To  find  the  opposite  side  a}  we  have  by  (1) 

a 

sin.  A  z=  T, 
h 

which,  multiplied  by  h,  gives 

a  =  h  sin.  A  ;  (17) 

or,  by  logarithms, 

log.  a  =  log.  h  +  log.  sin.  A. 
Thirdly.  To  find  the  side  6,  we  have  by  (4) 

cos.  A  =  -r, 
a 

which,  multiplied  by  h,  gives 

b  =  7i  cos.  .4  ;  (18) 

or,  by  logarithms, 

log.  b  =z  log.  7*  +  l°g*  cos.  ii. 


§  33.]  RIGHT    TRIANGLES.  21 

Leg  and  an  angle  given. 

32.  Problem.   To  solve  a  right  triangle,  when  a  leg 
and  the  opposite  angle  are  known.     [B.  p.  39.] 

Solution.  Given  (fig.  4.)  the  leg   a,  and  the  opposite  angle 
A,  to  solve  the  triangle. 

First.  The  angle  B  is  the  complement  of  A. 

Secondly.  To  find  the  hypothenuse  h,  we  have  by  (17) 
a  zz:  h  sin.  A, 
which,  divided  by  sin.  A,  gives  by  (6) 

h  =  ~ —  —  a  cosec.  A  ;  (19) 

sin.  A  v 

or,  by  logarithms, 

log.  h  =  log.  a  +  (ar.  co.)  log.  sin.  A 

s=  log.  a  -J-  log.  cosec.  A. 

Thirdly.     To  find  the  other  leg  6,  we  have  by  (4) 

b 

cotan.  .4  =  -, 

a 

which,  multiplied  by  a,  gives 

b  =i  a  cotan.  A  ;  (20) 

or,  by  logarithms, 

log.  b  =  log.  a  -j-  log.  cotan.  ^4. 

33.  Problem.    To  solve  a  right  triangle,  when  a  leg 
and  the  adjacent  angle  are  known.    [B.  p.  39.] 

Solution.  Given  (fig.  4.)  the  leg  a  and  the  angle  JB,  to  solve 
the  triangle. 

First.  The  angle  A  is  the  complement  of  B. 


22  PLANE  TRIGONOMETRY.  [CH.  III. 

Hypothenuse  and  a  leg  given. 

Secondly.  The  other  parts  may  be  found  by  (19)  and  (20), 
or  from  the  following  equations,  which  are  readily  deduced 
from  equations  (4)  and  (6),  ** 


h  z=  -  =  a  sec.  B, 

cos.   B 

(21) 

b  =z  a  tang.  B ; 

(22) 

or,  by  logarithms, 

log.  h  =z  log.  a  -f-  log.  sec.  B, 

log.  b  =  log.  a  +  log.  tang.  B. 

36.  Problem.   To  solve  a  right  triangle^  when  the 
hypothenuse  and  a  leg  are  known.    [B.  p.  40.] 

Solution.  Given  (fig.  4.)  the  hypothenuse  h  and  the  leg  a, 
to  solve  the  triangle. 

First.  The  angles  A  and  B  are  obtained  from  equation  (4), 

sin.  A  =  cos.  B  =  j ;  (23) 

or,  by  logarithms, 

log.  sin.  A  =  log.  cos.  B  =  log.  a  -j-  (ar.  co.)  log.  h. 

Secondly.  The  leg  b  is  deduced  from  the  Pythagorean  prop- 
erty of  the  right  triangle,  which  gives 

a2  +  62  =  h*,  (24) 

whence 

&2  p-  7,2  _  a2  =  (A  +  a)  (A  —  a), 
6  =  V  (7,2  _  a2)  —  ^  [(^  +  fl)  ^  _  a)]  j         (25) 
by  logarithms, 
log.  b.  =  i  log.  (A2—  a*)  =  J  [log.  (h  +  a)  +  log.  (A  — a)]. 


<§>  36.]  RIGHT    TRIANGLES.  23 

The  legs  given. 

35.  Problem.   To  solve  a  right  triangle,   when  the 
two  legs  are  known,    [B.  p.  40.] 

Solution.  Given  (fig.  4.)  the  legs  a  and   6,  to  solve  the 
triangle. 

First.  The  angles  are  obtained  from  (4) 

^        a 

tang.  A  =p  cotan.  B  =  - ; 

or,  by  logarithms, 

log.  tang.  A  =  log.  cotan.  B  =  log.  a  +  (ar.  co.)  log.  b. 

Secondly,  To  find  the  hypothenuse,  we  have  by  (24) 
h  =  V  (a2  +&2). 

Thirdly.  An  easier  way  of  finding  the   hypothenuse  is  to 
make  use  of  (19)  or  (20) 

h  =  a  cosec.  A  =  a  sec.  B ; 

or,  by  logarithms, 

log.  h  =  log.  a.  +  l°g-  cosec.  ^L  =  log.  a  -|-  log.  sec.  Z?. 


36.    Examples. 

1.  Given  the  hypothenuse  of  a  right  triangle  equal  to  49.58, 
and  one  of  the  acute  angles  equal  to  54°  44' ;  to  solve  the 
triangle. 

Solution.  The  other  angle  =  90°  —  54°  44'  =  35°  16'. 
Then  making  h  =  49.58,  and  A  =  54°  44';  we  have,  by  (17) 
and  (18), 


24  PLANE  TRIGONOMETRY.  [CH.  III. 

Examples  of  right  triangles. 

h  =  49. 58  1.69531  1.69531 

A  =  54°  44'     *  sin.  9.91194  cos.  9.76146 


a  =  40.  481  1.60725;  b  =  28. 627  1.45677. 

Am.  The  other  angle  =  35°  Iff; 

The  less -f    40'481> 
me  legs  _|    28.627. 

2.  Given  the  hypothenuse  of  a  right  triangle  equal  to  54.571, 
and  one  of  the  legs  equal  to  23.479  ;  to  solve  the  triangle. 

Solution.  Making  h  =  54.571,         a  ==  23.479; 
we  have,  by  (23), 


%      a  =  23.479 

1.37068 

h  =  54.571 

(ar.  co.)  8.26304 

A  =  25°  29'     sin. 
B  =  64°  31'     cos. 

|           9.63372. 

By  (25), 

h  +  a  =  78.050 

1.89237 

h  —  a  =  31.092 

1.49265 

b2 

2    |3.38502 

b  =  49.262 

1.69251 

Ans.     The  other  leg  =  49.262 

The  angles-}    f°0  JJ 

3.  Given  the  two  legs  of  a  right  triangle  equal  to  44.375, 
and  22.165;  to  solve  the  triangle. 

*  To  avoid  negative  characteristics  the  logarithms  are  retained 
as  in  the  tables,  according  to  the  usual  practice,  with  the  logarithms  of 
decimals,  as  in  B.  p.  29. 


<§>  36.]  RIGHT    TRIANGLES.  25 

Examples  of  right  triangles. 

Solution.  Making  a  —  44.375,  b  —  22.165  ;  we  have, 
a  —  44.375  1.64714  1.64714 

b  =  22.165       (ar.  co.)  8.65433 


A  z=  63°  27'  28"  tang.    )  1 0  om  47 .     cosec.  >    -  n  04fto7 
J5z=z26°32/32/cotan.  r030147'     sec.      j    1004837 

h  ==  49.603      1.69551 
Ans.     The  hypothenuse  ==  49.603 

{63°  °7/28// 
26^3232" 

4.  Given  the  hypothenuse  of  a  right  triangle  equal  to 
37.364,  and  one  of  the  acute  angles  equal  to  12°  30' ;  to  solve 
the  triangle. 

Ans.     The  other  angle  =  77°  30' 

rp,     ,  f    8.087 

The  legs  =  |  3a47g 

5.  Given  one  of  the  legs  of  a  right  triangle  equal  to  14.548, 
and  the  opposite  angle  equal  to  54°  24' ;  to  solve  the  triangle. 

Ans.     The  hypothenuse  ==  17.892 

The  other  leg—  10.415 

The  other  angle  =  35°  36^ 

6.  Given  one  of  the  legs  of  a  right  triangle  equal  to  11.111, 
and  the  adjacent  angle  equal  to  11°  ll7,  to  solve  the  triangle. 

Ans.      The  hypothenuse  =n  11.326 

The  other  leg  =z  2.197 

The  other  angle  =  78°  49;. 
3 


26  PLANE  TRIGONOMETRY.  [CH.  III. 


Examples  of  right  triangles. 


7.  Given  the  hypothenuse  of  a  right  triangle  equal  to  100, 
and  one  of  the  legs  equal  to  1,  to  solve  the  triangle. 

Arts.     The  other  leg  =  99.995 
0°  34'  23" 


The  angles  =  {  £  ^  23" 


8.  Given  the  two  legs  of  a  right  triangle  equal  to  8.148,  and 
10.864,  to  solve  the  triangle. 

Ans.     The  hypothenuse  =  13.58 
The  angle,  ={S:5?:$ 


§  38.]  GENERAL    FORMULAS.  27 

Sine  of  the  sum  of  two  angles. 


CHAPTER    IV. 


GENERAL    FORMULAS. 


37.  The  solution  of  oblique  triangles  requires  the 
introduction  of  several  trigonometrical  formulas,  which 
it  is  convenient  to  bring  together  and  investigate  all  at 
once. 

38.  Problem*  To  find  the  sine  of  the  sum  of  two 
angles. 

Solution.  Let  the  two  angles  be  BAC  and  B' AC  (fig.  9), 
represented  by  the  letters  M  and  N.  At  any  point  C,  in  the 
line  AC,  erect  the  perpendicular  BB1.  From  B  let  fall  on 
AB'  the  perpendicular  BP.  Then  represent  the  several  lines, 
as  follows, 

a  —  BC,  a1  =  BC,  b  =  AC 

h  =  AB,  h1  =  AB',  x  —  BP 

M=BAC,  NzzzBAC. 

Then,  by  (4), 

sin.  BAC  =  sin.  M  =  T,  sin.  N  z=z  %- 

ii  h; 

«r  b  b 

COS.  M =  T,  COS.  N  =  -77 

ft  /*' 

sin.  JBAP  =  sin.  (if  +  N)  =^-=|. 


28  PLANE  TRIGONOMETRY.  [CH.  IV. 

Sine  of  the  difference  of  two  angles. 

Now  the  triangles  BPB1  and  B'AC,   being  right-angled, 
and  having  the  angle  B1  common,  are  equiangular  and  similar. 

Whence  we  derive  the  proportion 

AB<  :  BB<  —  ACBP, 


whence 


and 


h!  :  a  +  « '  =  &  •  x  ? 

_ab  +  a'b 
X~         h'         ' 

,v,    .     ,T\        x      ab4-a'b 


The  second  member  of  this  equation  may  be  separated  into 
factors,  as  follows, 

/,r   .     ,Tx         ab  ,  ba' 


a     b  b     a' 


whence,  by  substitution,  we  obtain 

sin.  (i*f  +  N)  —  sin.  M  cos.  iV+  cos.  1/  sin.  iV.       (26) 

39.  Problem.  To  find  the  sine  of  the  difference  of 
two  angles. 

Solution.  Let  the  two  angles  be  BAC  and  B 'AC  (fig.  10), 
represented  by  M  and  N.  At  any  point  C  in  the  line  AC 
erect  the  perpendicular  BBC.  From  B  let  fall  on  AB'  the 
perpendicular  JBP.  Then,  using  the  notation  of  §  38,  we 
have 

sin.  BAP  ==  sin.  (M—  N)  =  ^  —  | 

7       A  B        h 


$  40.]  GENERAL    FORMULAS.  29 

Cosine  of  the  sum  of  two  angles. 

The  triangles  B' AC  and  BB'P  are  similar,  because  they 
are  right-angled,  and  the  angles  at  B'  are   vertical   and  equal. 

Whence 


AB1  :  BB'  —  AC:  BP, 

or 

h'  :  a  —  a1  =  b  :  x  ; 

whence 

ab  —  a'b 
X  =         A'"' 

and 

sin. 

, ,.  _        mr%         x         «  b  —  ba' 

(*~^=;A=        ft    - 

=  p'  —  TV 
a     b         b     a! 

and  by  substitution, 

sin.  (M —  N)  =  sin.  M  cos.  iV —  cos.  M  sin.  iV.        (27) 

40.  Problem.   To  find  the  cosine  of  the  sum  of  two 
angles. 

Solution.  Making  use  of  (fig.  9),  with  the  notation  of  §  38, 
and  also  the  following 

!/  =  AP,z=PB>; 

we  have 

co,(^  +  iV)=^|=f- 

But 

y  =  AB'  —  PB<  =zh'  —  z. 
3* 


30  PLANE  TRIGONOMETRY.  [cH.  it. 

Cosine  of  the  sum  of  two  angles. 

The  similar  triangles  BPB'  and  B'AC,  give  the  propor- 
tion 

AB'  :  BB'  sb  BC  .  B'P} 


h1  :  a  -f-  a'  =  a1  :  z; 

whence 

«  a'  +  a'  2 

*  =  — ¥~ ; 

and 

y  —  h'  —  zzzzh' j± 

_  &/2  __  a/2        rt  ^ 
~~  ^  " 

But,  from  the  right  triangle  AB'C, 

h'2  —  a'2  ss  (4JB»)  2  —  (^'C)2  =b  (4C)2  bb  &2  ; 
whence 


and 


y 

_  b2 

h> 

«  «' 

n  + 

iV)  = 

.y 
'  h 

b2 

—  ad 
hh' 

_  h2 
~hh' 

— 

aa! 
W 

b 

~ h 

b 

a 

a' 

whence  by  substitution, 

cos.  (M+  N)  =z  cos.  M.  cos.  iV —  sin.  M.  sin.  JV.      (28) 

41.  Problem.   To  find  the  cosine  of  the  difference  of 
tioo  angles. 


§  41.]  GENERAL    FORMULAS.  31 

Cosine  of  the  difference  of  two  angles. 

Solution.  Making  use  of  (fig.  10.)  with  the  notation  of  the 
preceding  section,  we  have 

cos.  BAB'  ==  cos.  (M—  N)  =  ^-=  =  f. 
v  '        AB        h 

But  y  —  AB'  +  PB'  =  h'  +  ft 

The  similar  triangles  BB'P  and  B AC  give  the  proportion 
.4J3'  :  BB'  —  B'C :  J3P, 

or  h'  :  a  —  d  =  d  :  z ; 


whence 
and 

a  d  —  a'2 

_h'2  —  a'2  +  ad 

h' 

But 

h'2  —  a12  =  b2. 

Hence 

b2  -f-  ad 

and         cos. 

<*T?(H=^ 

62     ,     aa' 

b     b         a     d 

or,  by  substitution, 

cos.  (if  —  N)  =  cos.  if  cos.  iV+  sin.  if  sin.  iV.        (29) 


32  PLANE  TRIGONOMETRY.  [CH.  IV. 

Sum  and  difference  of  sines  and  cosines. 

42.  Corollary.  The  similarity,  in  all  but  the  signs,  of  the 
formulas  (26)  and  (27)  is  such,  that  they  may  both  be  written 
in  the  same  form,  as  follows, 

sin.  (Jf  dc  N)  =  sin.  M  cos.  N  =b  cos.  M  sin.  N.         (30) 

in  which  the  upper  signs  correspond  with  each  other,  and  also 
the  lower  ones. 

In  the  same  way,  by  the  comparison  of  (28)  and  (29),  we 
are  led  to  the  form 

cos.  (Jf  dc  N)  =s  cos.  Jf.  cos.  N  ^F  sin.  Jf.  sin.  N,      (31) 

in  which  the  upper  signs  correspond  with  each  other,  and  also 
the  lower  ones. 


43.  Corollary.  The  sum  of  the  equations   (26)  and  (27)  is 
sin.  (Jf  +  N)  +  sin.  (M  —  N)  =  2  sin.  31  cos.  N.    (32) 

Their  difference  is 

sin.  (Jf  +  N)  —  sin.  (Jf  —  N)  =  2  cos.  M  sin.  N.    (33) 

44.  Corollary.  The  sum  of  (28)  and  (29)  is 

cos.  (Jf  +  N)  +  cos.  (Jf  —  N)  =  2  cos.  Jf  cos.  iV.    (34) 
Their  difference  is 

cos.  (Jf  —  N)  —  cos.  (Jf  +  2V)  =  2  sin.  Jf  sin.  JV.    (35) 

45.  Corollary.  If,  in  (32-25),  we  make 

M+N=  A,  andM—N=:B; 

that  is, 

3r=i(A  +  B),  N  =  i  (A  —  B) ; 

they  become,  as  follows, 


<§>  48.]  GENERAL    FORMULAS.  33 


Sum  and  difference  of  sines  and  cosines. 

sin.  A  +  sin.  B  =  2  sin.  £  (A  +  #)  cos.  J  (^L  — JB)  (36) 

sin.  A  —  sin.  JB  =  2  cos.  £  (it  -f  B)  sin.  £  (4  —  B)  (37) 

cos.  ^4  +  cos.  B  =  2  cos.  £  (4  +  JB)  cos.  |  (.4  —  J3)  (38) 

cos.  B  —  cos.  A  =2  sin.  J  (4  +  5)  sin.  |  (4  —  jB).  (39) 

46.  Corollary.  The  quotient,  obtained    by  dividing  (36)  by 
(37),  is 

sin.  A  +  sin.  B  __  sin.  £  (A -{- B)  cos.  ±  (A  —  B) 
sin.  ^4  —  sin.  B         cos.  ±  (A  -\-  B)  sin.  £  (A  —  B)' 

Reducing  the  second  member  by  means  of  equations  (6),  (7), 
(8),  we  have 

sin.  A  4-  sin.  B  «-«-*»*  ,  -         « 

sTnTZ^smTl*  *  tang'  *  (^  +  *)  COtan'  i(A~B) 

j  tang.^+#)  =  cotan.  j.  (.4  —  _2J) 

tang.  £  (4  —  #)         cotan.  $  (A  +  £)'  *     ' 

47.  Corollary.  The  quotient  of  (39)  divided  by  (38)  is,  by 
reduction, 

cos.  B  —  cos.  A  .j  ,\  \  ■  '  _  %  „  ^     •    • 

=  tang.^  +  I?)  __  tang.j(il  — g 

cotan.  £(A—B)        cot*n.£(A+ B'  {     ' 

48.  Corollary.  Putting  in   (26)  and    (28),  M  and   iV  both 
equal  to  Ai  we  obtain 

sin.  2  A=  sin.  ^4  cos.  A -\- sin.  A  cos.  A  =  2  sin.  ^4  cos.  A  (42) 

cos.  2  ^4  =  cos.  A  cos.  ^4  —  sin.  A  sin.  ^4 

=  (cos.  A)*  —  (sin.  A)*.  (43) 


34  PLANE  TRIGONOMETRY.  [CH.  IV. 

Sine  &c.  of  double,  and  half  of  an  angle. 

49.  Corollary.   The  sum    of  (43),  and    of  the   following 
equation,  which  is  the  same  as  (9), 

1  =  (cos.  A)2  -f  (sin.  A)2, 

is  1  +  cos.  2  A  =  2  (cos.  A)2.  (44) 

Their  difference  is 

1  —  cos.  2  A  =  2  (sin.  A)2.  (45) 

50.  Corollary.  Making  2  A  =  C,  or  C=  £  A,  in  (42-49), 
we  obtain 

sin.  C  =  2  sin  £  C  cos.  J  C  (46) 

cos.  C  =  (cos.  J  C)2  —  (sin.  J  C)*  (47) 

1  +  cos.  C  =  2  (cos.  J  C)2  (48) 

1  _  cos.  C  =  2  (sin.  J  C)2.  (49) 
The  equations  (48)  and  (49)  give 

cos.  i  C  ==  V[i  (1  +  cos.  C)]  (50) 

sin.  J  C  =  V[J(1  —  cos.  C)]  (51) 

-.      ^  #/l    COS.   C\  ,__, 

51.  Problem.    To  find  the  tangent  of  the  sum  a?id  of 
the  difference  of  two  angles. 

Solution.    First.    To  find  the  tangent  of  the   sum  of  two 
angles,  which  we  will  suppose  to  be  M  and  N9  we  have  from 

(*% 

tmm    ,     ™         s\n.  (M+N) 

tang.  (M  +  N)  = W-T   »n'- 

b   v      ~      '        cos.  (if  +  iV) 


§   51,]  GENERAL    FORMULAS.  35 


Tangent  of  sum  and  difference  of  angles. 


Substituting  (26)  and  (28), 

sin.  Mcos.  N-\-  cos.  M sin.  N 


tang.  (M-{~  N)  = 


cos.  Mcos.  N —  sin.  M sin.  N' 


Divide  every  term  of  both  numerator  and  denominator  of  the 
second  member  by  cos.  31  cos.  N ; 

sin.  M  cos.  N        cos.  ill"  sin.  N 

I  ht   .     7%rv         cos-  M  cos.  iV     ■"  COS.  M  cos.  N 

tang.  (M-h-  N)  = — , 

&    v        '        '         cos.  M  cos.  iV         sin.  if  sin.  N 

cos.  Jf  cos.  iV         cos.  M  cos.  iV 

sin.  M  sin.  iV 

_  cos.  M     '      cos.  iV 
"  ~        sin.  M  sin.  iV' 
cos.  iH '  cos.  iV 

which,  reduced  by  means  of  (7),  becomes 

mr  \     7»rv        tang.  Jf  4- tang.  iV 
tang.  (Jf+  2V)  =  -^—t-l-.  (68) 

Secondly.    To  find  the  tangent  of  the  difference  of  M  and 
N,  since  by  (7) 

tang.  (Jf-jy)  =  8in- <*-*[) 
v  '       cos.  (M—  JV)9 

a  bare  inspection  of  (30)  and  (31)  shows  that  we  have  only 
to  change  the  signs,  which  connect  the  terms  in  the  value  of 
tang.  (M  +  N)  to  obtain  that  of  tang.  ( M  —  N).  This  change, 
being  made  in  (53),  produces 

•tang.  (M -  N)  =  .**ng.  if- tang.  *T 

S  V  ;       1  +  tang.  M  tang.  JV  l     > 


36  PLANE  TRIGONOMETRY.  [CH.  IV. 


Tangent  and  cotangent  of  double  an  angle. 


52.  Corollary,  As  the  cotangent  is  merely  the  reciprocal  of 
the  tangent,  we  have,  by  inverting  the  fractions,  from  (53)  and 
(54), 

cotan.  (M+N)  = i#-r~: ^-*r,  (55) 

v        '        '          tang.  M  -f-  tang.  N  '  v      ' 

1  +  tan£-  ^"tano*.  N  ,   " 

cotan.  (M—N)z=  —X — =1 2—.  (56) 

v               t          tang.  JIT — tang.  N  v     ' 

53.  Corollary.  Make  M  =  X  =  A,  in  (53)  and  (55). 
They  become 

_    .               2  tang.  A  ,^v 

cotan^^1-^"^.  (58) 

2  tang,  vl  v     ' 


§  54.]  PARTICULAR  VALUES  OF  SINES,  &C.  37 

Sine,  &c.  of  0°  and  90°. 


CHAPTER   V. 

VALUES    OF  THE    SINES,  COSINES,    TANGENTS,    COTANGENTS, 
SECANTS,  AND  COSECANTS  OF   CERTAIN  ANGLES. 

54.  Problem.   To  find  the  sine,  fyc.  of0°  and  90°. 

Solution.  Supposing  M  =  N,  in  (27)  and  (29),  we  have 

sin.  (M —  M)  =  sin.  0°  ==  sin.  M  cos.  M —  cos.  iJfsin.  M 

cos.  (M—M)  =  cos.0°  —  (cos.  M)2  +  (s'm.M)2  ; 

whence,  by  (9),   and  the   consideration  that  0°  and  90°  are 
complements  of  each  other, 

sin.  0°  —  cos.  90°  =  0  (59) 

cos.  0°  =  sin.  90°  =  1.  (60) 

From  (6)  and  (7),  we  have 

tang.  0°  sa  cotan.  90°  =  Sm'  °no  =  °  =  0,  (61 ) 

cos.  0  *  v     y 

cotan.  0°  =  tang.  90°  =  — -— b  =  4-  =  oo         (62) 
tang.  0  x     ' 

sec.  0°  =  cosec.  90°  = —  =  4.  =  I  (63) 

cos.  0 

cosec.  0°  =  sec.  90°  =  - — -*■  =  i  =  oo.  (64) 

sm.  0  u 


38  PLANE  TRIGONOMETRY.  [CH.  V. 

Sine,  &c.  of  180°  and  270". 

55.  Problem.   To  find  the  sine,  $*c.  of  180°. 

Solution.  Make  A  ==  90°,  in  (42)   and  (43),  they  become, 
by  means  of  (59)  and  (60), 

sin.  180°  =  2  sin.  90°  cos.  90°  =  0  (65) 

cos.  180°  ===  (cos.  90°)2  —  (sin.  90°)2  =  —  1.       (66) 

Hence  from  (6)  and  (7), 

n     1ftO°  n 

=  0  (67) 


=  —  oo        (68) 
=  -  1        (69) 


tang. 

TSOo  _  sin.  180° 
180    ~  cos.  180° 

0 
—  1 

cotan. 

180o  =  cos.  180' 
sin.  180° 

—  1 

sec. 

180°  — 

i 

~~  cos.  180° 

—  i 

cosec. 

180°     -  Sin'  18°° 

=    IT    = 

(70) 


56.  Problem.  To  find  the  sine,  Sfc.  of  270°. 


Solution.  Make  M  ==  180°  and  N  =  90°  in  (26)  and  (28). 
They  become,  by  means  of  (59,  60,  6o}  66), 

sin.  270°  =  sin.  180°  cos.  90°+cos.  180°  sin.  90°=  — 1  (71) 

cos.  270°  =  cos.  180°  cos.  90°  —  sin.  180°  sin.  90°  =  0.  (72) 

Hence,  from  (6)  and  (7), 

0_0        sin.  270°        —  I  - 

tang-270  =c^27(F==-(r  =  ~w    (73) 

«^o        cos.  270°  0 

COtan"  27°    =  »1F  =  —1  =  °  <74> 


§  58.]  PARTICULAR  VALUES  OF  SINES,  &C.  39 

Sine,  &c.  of  360°  and  45°. 

:  .»o-2w=s^W=*=-      <75> 


57.  Problem.    To  find  the  sine,  fyc.  of  360°. 

Solution.  Make  A  =  180°  in  (42)  and  (43) ;  and  they  be- 
come by  (65,  66,  59,  60) 

sin.  3G0°  =  0  =  sin.  0°  (77) 

cos.  360°  cs=  1  =  cos.  0°.  (78) 

Hence  the  sine,  fyc.  of  360°  are  the  same  as  those  of0°. 

58.  Problem.    To  find  the  sine,  fyc.  of  45°. 

Solution.  Make  C  =  90°  in  (50)  and   (51).  They  become, 
by  means  of  (59), 

cos.  45°  =  V  [J  (1  +  cos.  90°)]  z=z  «/i  (79) 

sin.  45°  —  s/  [J  (1  —  cos.  90°)]  = +/i  =  cos.  45°.   (80) 

Hence,  from  (6)  and  (7), 

tang.  45>  ==  "  =  1  (81) 

cos.  45  v     ' 

cotan.  45°  = -_  =  1  =  tang.  45°  (82) 

tang.  45  to  v     / 

sec.  45°  = X—  =  -±--^2  (83) 

cos.  45  \/  £  v     ' 

cosec.  45    = 


-r— -0-  =  — - -  =  s/2  =  sec.  45°.     (84) 
sin.  45°         s/  £  v     ' 


40  PLANE  TRIGONOMETRY.  [CH.  V. 

Sine,  &c.  of  30°,  60^,  and  the  supplement. 

59.  Problem.   To  find  the  sine,  fyc.  of  30°  and  60°. 

Solution.  Make  A  =  30°  in  (42).  It  becomes,  from  the 
consideration  that  30°  and  60°  are  complements  of  each 
other, 

sin.  60°  =  cos.  30°  =-  2  sin.  30°  cos.  30°. 

Dividing  by  cos.  30°,  we  have 

1  =  2  sin.  30°, 

or  sin.  30°  =  J  =  cos.  60°  (85) 

whence,  from  (6),  (7),  and  (10), 

cos.  30°  =  sin.  60°  =±  s/  (1  —  \)  =  \  \/3  (86) 

tang.  30°  =£  cotan.  60°  ==  j-^r-  =  ^-  =  Vi  (87) 

cotan.  30°  =  tang.  60°  =z  -j—  =  \/3  (88) 

1  2 

sec.  30°  =  cosec.  60°  =  - — —  =  -7--  (89) 

^  V  o        \/  o 

cosec.  30°  ==  sec.  60°  =  -  =  2.  (90) 

60.  Problem.  To  find  the  sine,  Sfc.  of  the  supplement 
of  an  angle. 

Solution.  Make  M  =  180°  in  (27)  and  (29).  They  become, 
by  means  of  (65)  and  (66), 

sin.  (180°  —  2V}==  sin.  180°  cos.  N  —  cos.  180°  sin.  N 

=  sin.  iV  (91) 

cos.  (180°  —  N)z=z  cos.  180°  cos.  JY  +  sin.  180°  sin.  iV 
==  —  cos.  JV  (92) 


$  62.]  PARTICULAR  VALUES  OF  SINES,  &C.  41 

Sine,  <&c.  of  obtuse  angle. 

whence,  from  (6)  and  (7), 

tang.  (180°  —  N)  =  —  tang.  N  (93) 

cotan.  (180°  —  N)  =  —  cotan.  N  (94) 

sec.  (180°  —  N)  —  —  sec.  N  (95) 

cosec.  (180°  —  N)  =  cosec.  N;  (96) 

that  is,  the  sine  and  cosecant  of  the  supplement  of  an 
angle  are  the  same  with  those  of  the  angle  itself  and 
the  cosine,  tangent,  cotangent,  and  secant  of  the  supple- 
ment are  the  negative  of  those  of  the  angle. 

61.  Corollary.  Since,  when  an  angle  is  acute  its 
supplement  is  obtuse,  it  follows  from  the  preceding 
proposition,  that  the  sine  and  cosecant  of  an  obtuse 
angle  are  positive,  while  its  cosine,  tangent,  cotangent, 
and  secant  are  negative. 

This  proposition  must  be  carefully  borne  in  mind  in  using 
the  trigonometric  tables,  as  it  affords  the  means  of  discrimi- 
nating between  the  two  angles  which  are  given  in  B.  Table 
XXVII,  and  of  deciding  which  of  these  two  angles  is  the 
required  one. 

62.  Corollary.  The  preceding  corollary  might  also  have 
been  obtained  from  (26)  and  (28).  For  by  making  M  —  90°, 
we  have  by  (59)  and  (60) 

sin,  (90°  +  N)  =  cos.  N  (97) 

cos.  (90°  +  N)  =  —  sin.  N\  (98) 

whence,  by  (6)  and  (7), 

tang.  (90°  +  N)  =  —  cotan.  N  (99) 

4* 


42  PLANE  TRIGONOMETRY.  [CH.  V. 

Sine,  &c.  of  negative  angle. 

cotan.  (90°  +  N)  =  —  tang.  N  (100) 

sec.  (90°  +  N)  =  —  cosec.  iV  (101) 

cosec.  (90°  +  N)  =  sec.  iV;  (102) 

that  is,  the  sine  and  cosecant  of  an  angle,  which  exceeds 
90°,  are  equal  to  the  cosine  and  secant  of  its  excess 
above  90°7  while  its  cosine,  tangent,  cotangent,  and  se- 
cant are  equal  to  the  negative  of  the  sine,  cotangent, 
tangent,  and  cosecant  of  this  excess. 

63.  Problem.    To  find  the  sine,  &c.   of  a  negative 
angle. 

Solution,  Make  N  =  0°  in  (27)  and  (29).     They  become, 
by  means  of  (59)  and  (60), 

sin.  (  11  N)  =  —  sin.  N  (103) 

cos.  (-N)  —cos.N  (104) 

whence,  from  (6)  and  (7), 

tang.  (  —  N)  —  —  tang.  N  (105) 

cotan.  (  — -  N)  =  —  cotan.  N  (106) 

sec.  (  —  N)  =  sec.  N  (107) 

cosec.  (  —  N)  —  —  cosec.  N;  (108) 

so  that  the  cosine  and  secant  of  the  negative  of  an 
angle  are  the  same  with  those  of  the  angle  itself ;  and 
the  sine,  tangent,  cotangent,  arid  cosecant  of  the  nega- 
tive of  the  angle  are  the  negative  of  those  of  the 
angle. 


§  66.]  PARTICULAR  VALUES  OF  SINES,  &C.  43 

Sine,  &c.  of  an  angle  greater  than  180°. 

64.  Problem,  To  find  the  sine,  Sfc.  of  an  angle 
lohich  exceeds  180°. 

Solution.  Make  M  —  180°  in  (26)  and  (28).  They  be- 
come, by  means  of  (65)  and  (66), 

sin.  (180°  +  N)  =  —  sin.  N  (109) 

cos.  (180°  +  N)  =z  —  cos.  N  (110) 

whence,  from  (6)  and  (7), 

tang.  (180°  +  N)  =  tang.  N  (111) 

cotan.  (180°  +  N)  =  cotan.  N  (112) 

sec.  (180°  +  N)  =  —  sec.  N  (113) 

cosec.  (180°  +  N)  =  —  cosec.  2V;  (114) 

that  is,  the  tangent  and  cotangent  of  an  angle,  which 
exceeds  180°,  are  equal  to  those  of  its  excess  above  180° ; 
and  the  sine,  cosine,  secant,  and  cosecant  of  this  angle 
are  the  negative  of  those  of  its  excess. 

65.  Corollary.  If  the  excess  of  the  angle  above  180° 
is  less  than  90°,  the  angle  is  contained  between  180° 
and  270° ;  so  that  the  tangent  and  cotangent  of  an 
angle  which  exceeds  180°,  and  is  less  than  270°,  are 
positive  ;  while  its  sine,  cosine,  secant,  and  cosecant  are 
negative. 

66.  Corollary*  If  the  excess  of  the  angle  above  180° 
is  greater  than  90°  and  less  than  180°,  the  angle  is 
contained  between  270°  and  360°;  so  that,  by  §  64  and 
61,  the  cosine  and  secant  of  an  angle,  which  exceeds 


44  PLANE  TRIGONOMETRY.  [CH.  V. 


270°  and  is  less  than  360°,  is  positive  ;  while  its  sine, 
tangent,  cotangent,  and  cosecant  are  negative. 

67.  Corollary.  The  results  of  the  two  preceding  corollaries 
might  have  been  obtained  from  (27)  and  (29).  For  by  mak- 
ing M  =  360°,  we  have,  by  §  57, 

sin.  (360°  —  N)  =  —  sin.  N  (115) 

cos.  (360°  —  N)—  cos.  N  (116) 

whence,  by  (6)  and  (7), 

tang.  (360°  —  N)  =  —  tang.  N  (117) 

cotan.(360°  —  N)  =  —  cotan.  N  (118) 

sec.  (360°  —  N)=z  sec.  N  (119) 

cosec.  (360°  —  N)  =  —  cosec.  N  (120) 

that  is,  the  cosine  and  secant  of  an  angle  are  the  same 
with  those  of  the  remainder  after  subtracting  the  angle 
from  360° ;  while  its  sine,  tangent,  cotangent,  and  co- 
secant are  the  negative  of  those  of  this  remainder* 

68.  Problem.  To  find  the  sine,  fyc.  of  an  angle  which 
exceeds  360°. 

Solution.  Make  M  =  360°  in  (26)  and  (28;.  They  be- 
come, by  means  of  (77)  and  (78), 

sin.  (360°  -f  N)  =  sin.  N  (121) 

cos.  (360°  +  N)  =  cos.  JY  (122) 

that  is,  the  sine,  fyc.  of  an  angle  which  exceeds  360° 
are  equal  to  those  of  its  excess  above  360°. 


<§>  70.]  PARTICULAR  VALUES  OF  SINES,  &C.  45 

Increase  of  sine,  &c.  of  an  acute  angle. 

69.  Theorem.  The  sine,  tangent,  and  secant  of  an 
acute  angle  increase  with  the  increase  of  the  angle ; 
the  cosine,  cotangent,  and  cosecant  decrease. 

Proof.  I.  The  excess  of  the  sine  of  M  -|-  m  over  the  sine 
of  M  is,  by  (13),  equal  to  sin.  m  cos.  M}  which  is  a  positive 
quantity  when  31  is  acute ;  and,  therefore,  the  sine  of  the  acute 
angle  increases  with  the  increase  of  the  angle. 

II.  The  excess  of  cos.  M  over  cos.  (M  -{-  m)  is,  by  (15), 
equal  to  sin.  m  sin.  M,  which  is  a  positive  quantity  ;  and, 
therefore,  the  cosine  of  the  acute  angle  decreases  with  the 
increase  of  the  angle. 

III.  The  tangent  of  an  angle  is,  by  (7),  the  quotient  of  its 
sine  divided  by  its  cosine.  It  is,  therefore,  a  fraction  whose 
numerator  increases  with  the  increase  of  the  angle,  while  its 
denominator  decreases.  Either  of  these  changes  in  the  terms 
of  the  fraction  would  increase  its  value ;  and,  therefore,  the 
tangent  of  an  acute  angle  increases  with  the  increase  of  the 
angle. 

IV.  The  cosecant,  secant,  and  cotangent  of  an  angle  are, 
by  (6),  the  respective  reciprocals  of  the  sine,  cosine,  and  tan- 
gent. But  the  reciprocal  of  a  quantity  increases  with  the 
decrease  of  the  quantity,  and  the  reverse.  It  follows,  then, 
from  the  preceding  demonstrations,  that  its  secant  increases 
with  the  increase  of  the  acute  angle,  while  its  cosecant  and 
cotangent  decrease. 

70.  Theorem.  The  absolute  values  (neglecting  their 
signs)  of  the  sine,    tangent,  and   secant  of  an  obtuse 


46  PLANE  TRIGONOMETRY.  [CH.  V. 

Increase  of  sine,  &c.  of  obtuse  angle. 

angle  decrease  with  the  increase  of  the  angle;  while 
those  of  the  cosine,  cotangent,  and  cosecant  increase. 

Proof.  The  supplement  of  an  obtuse  angle  is  an  acute 
angle,  of  which  the  absolute  values  of  the  sine,  &,c.  are,  by  § 
60,  the  same  as  those  of  the  angle  itself.  But  this  acute  angle 
decreases  with  the  increase  of  the  obtuse  angle,  and  at  the 
same  time  its  sine,  tangent,  and  secant  decrease,  while  its 
cosine,  cotangent,  and  cosecant  increase. 


§  71.]  OBLIQUE    TRIANGLES.  47 

Sides  proportional  to  sines  of  opposite  angles. 


CHAPTER    VI. 

OBLIQUE    TRIANGLES. 

71.  Theorem.  The  sides  of  a  triangle  are  directly 
proportional  to  the  sines  of  the  opposite  angles.  [B.  p.  13] 

Proof.  In  the  triangle  ABC  (figs.  2  and  3),  denote  the 
sides  opposite  the  angles  A,  B,  C,  respectively,  by  the  letters 
a,  by  c.     We  are  to  prove  that 

sin.  A  :  sin.  B  :  sin.  C  —  a  :b  :  c.  (123) 

From  the  vertex  B,  let  fall  on  the  opposite  side  the  perpendic- 
ular BP,  which  we  will  denote  by  the  letter  p.  Then,  in  the 
triangle  BAP,  we  have  by  (1) 

Sm'A  =  AB:=-c' 
or  p  =zc  sin.  A.  0^4) 

Also,  in  the  triangle  BPC,  we  have,  by  (1)  and  (91),  and 
from  the  consideration  that  BCP  is  the  angle  C  (fig.  2.),  and 
its  supplement  (fig.  3.), 


BCP  = 


BP       p 


~~  BC~  a' 

or  p  =za  sin.  C.  (125) 

Comparing  (124)  and  (125),  we  have 

c  sin.  A  =  a  sin.  C, 


48  PLANE  TRIGONOMETRY.  [CH.  VI. 

A  side  and  two  angles  given. 

which  may  be  converted  into  the  following  proportion 

sin.  A  :  sin.  C  z=  a  :  c. 
In  the  same  way,  it  may  be  proved  that 

sin.  A  :  sin.  B  —  a  :  b  ; 
and  these  two  proportions  may  be  written  in  one  as  in  (123). 

72.  Problem.  To  solve  a  triangle  when  one  of  its 
sides  and  two  of  its  angles  are  known.   [B.  p.  41.] 

Solution.  First.  The  third  angle  may  be  found  by  subtract- 
ing the  sum  of  the  two  given  angles  from  180°. 

Secondly.  To  find  either  of  the  other  sides,  we  have  only  to 
make  use  of  a  proportion,  derived  from  §  71.  As  the  sine  of 
the  angle  opposite  the  given  side  is  to  the  sine  of  the  angle 
opposite  the  required  side,  so  is  the  given  side  to  the  required 
side.  Thus,  if  a  (fig.  1.)  were  the  given  and  b  the  required 
side,  we  should  have  the  proportion 

sin.  A  :  sin.  B  —  a  :  b  ; 

whence  by  (6) 

a  sin.  B  .      _  «-^l» 

b  —  — : --  ±=  a  sin.  B  cosec.  A.  (126) 

sin.  A  x       ' 


73.    Examples. 

1.  Given  one  side  of  a  triangle  equal  to  22.791,  and  the 
adjacent  angles  equal  to  32°  41/  and  47°  54' ;  to  solve  the 
triangle. 

Solution.  The  other  angle  •=  180°  —  (32°  417  +  47°  54') 

=  99°  257. 


*74.] 


OBLIQUE    TRIANGLES. 


49 


Given  two  sides  and  an  angle  opposite  one  of  them. 


By  (126) 


99°  25'  cosec. 

10.00589 

10.00589 

32°  41'  sin. 

9.73239 

47°  54'  sin. 

9.87039 

22.791 

1.35776 

1.35776 

12.475 


*1.09604;         17.141         *i.  23404. 
Ans.  The  other  angle  ==  99°  25' 


The  other  sides 


=  { 


12.475 
17.141 


2.  Given  one  side  of  a  triangle  equal  to  327.06,  and  the 
adjacent  angles  equal  to  154°  22'  and  17°  35' ;  to  solve  the 
triangle. 

Ans.  The  other  angle  =  8°  3' 


The  other  sides 


-{ 


1010.4 
705.5 


74.  Problem.  To  solve  a  triangle  when  two  of  its 
sides  and  an  angle  opposite  one  of  the  given  sides  are 
known.  [B.  p.  42.] 

Solution.  First.  The  angle  opposite  the  other  given  side  is 
found  by  the  proportion  of  §  71.  As  the  side  opposite  the 
given  angle  is  to  the  other  given  side,  so  is  the  sine  of  the 
given  angle  to  the  sine  of  the  required  angle.  Thus,  if  (fig.  1.) 
a  and  b  are  the  given  sides  and  A  the  given  angle,  the  angle 
B  is  found,  by  the  proportion 


*  20  is  subtracted  from  each  of  these  characteristics,  because  the 
two  sines  and  cosecant  were  taken  from  the  tables  without  any  di- 
minution, as  required  by  §  30. 

5 


50  PLANE  TRIGONOMETRY.  [CH.  VI. 

Given  two  sides  and  an  angle  opposite  one  of  them. 

a  \b  =.  sin.  A  :  sin.  B ; 
whence 

S\n.B=bsmaA.  (127) 

Secondly.  The  third  angle  is  found  by  subtracting  the  sum 
of  the  two  known  angles  from  180°. 

Thirdly.  The  third  side  is  found  by  the  proportion.  As  the 
sine  of  the  given  angle  is  to  the  sine  of  the  angle  opposite  the 
required  side,  so  is  the  side  opposite  the  given  angle  to  the 
required  side.     That  is,  in  the  present  case, 

sin.  A  :  sin.  C  =  a  :  c  ; 
whence 

,  —  a  sin.  c  cosec.  A.  (128) 


sin.  A 


75.  Scholium.  Two  angles  are  given  in  the  tables  corre- 
sponding to  the  same  sine,  which  are  supplements  of  each 
other,  one  being  acute  and  the  other  obtuse.  Two  values  of 
B  (127)  are  then  given  in  the  tables,  and  both  these  values 
may  be  possible,  when  the  given  value  of  b  is  greater  than  that 
of  a>  and  the  given  value  of  A  is  less  than  90°  ;  for,  in  this 
case,  there  may  be  two  triangles,  ABC  (fig.  11.)  and  AB'C, 
which  satisfy  the  data. 

76.  Scholium.  The  problem  is  impossible,  when  the  given 
value  of  b  is  greater  than  that  of  a,  and  the  given  value  of  A 
is  obtuse.  For  the  greatest  side  of  an  obtuse-angled  triangle 
must  always  be  opposite  the  obtuse  angle. 

77.  Scholium.  The  problem  is  impossible,  when  the  given 
value  of  b  is  so  much  greater  than  that  of  a,  that  we  have 


§   79.]  OBLIQ,UE    TRIANGLES.  51 

Given  two  sides  and  an  angte  opposite  one  of  them. 

b  sin.  A  >  a; 

for,  in  this  case,  the  given  value  of  a  is  less  than  that  of  the 
perpendicular  CP  (fig.  11.),  from  C  upon  AP. 

78.  Scholium.  The  obtuse  value  of  B  does  not  satisfy  the 
problem,  when  b  is  less  than  a;  for  the  obtuse  angle  of  a 
triangle  cannot  be  opposite  a  smaller  side.  In  this  case, 
therefore,  the  problem  admits  of  only  one  solution. 


79.    Examples. 

1.  Given  two  sides  of  a  triangle  equal  to  77.245  and  92.341, 
and  the  angle  opposite  the  first  side  equal  to  55°  28'  12",  to 
solve  the  triangle. 

Solution.    Making 

b  —  92-341,  a  =  77-245,  A  =  55°  28'  12", 

we  have,  by  (127), 

a  =  77-245  (ar.  co.)         8-11213 

b   =  92-341  1-96540 

A  =  55°  28;  12"  sin.     991584 


B  =  80;  1'  or  =  99°  59'  sin.     9-99337 

A  +  B  =  135°  29'  12"  or  ==  155°  27;  12" 
C  =    44°  30'  48"  or  =    24°  32'  48" 
Then,  by  (128), 


52  PLANE  TRIGONOMETRY.  [CH.  VI. 

Given  two  sides  and  an  angle  opposite  one  of  them. 


a  =  77-245  1-88787  1-88787 

C  =  44°  30' 48"    sin.    9-84576  or  ===  24° 32' 48" sin.  961850 
A  ==  55°28  12"cosec.  1008416  1008416 


C=  65-734  1-81779  or  =  38  952  1-59053 

Ans.  The  third  side  =     65-734         or  =     38-952 

{80°  V  i  99°  59' 

44°  30' 48"  ov  —  {  24°  32'  48" 

2.  Given  two  sides  of  a  triangle  equal  to  77-245  and  92-341, 
and  the  angle  opposite  the  second  side  equal  to  55°  28'  12"  ; 
to  solve  the  triangle. 

Ans.  The  third  side       =  110-7 

T,       4,  ,  (   43°  3344" 

The  other  angles    =    j   g()0  5g,    4„ 

3.  Given  two  sides  of  a  triangle  equal  to  40  and  50,  and  the 
angle  opposite  the  first  side  equal  to  45°,  to  solve  the  tri- 
angle. 

Ans.  The  third  side  =     54.061        or  =     16-65 


{62°     7'  ( 

72°  W       or  =  i 


117°  53' 

17°  r 


4.  Given  two  sides  of  a  triangle  equal  to  77-245  and  92*341, 
and  the  angle  opposite  the  second  side  equal  to  124°  31'  48", 
to  solve  the  triangle. 

Ans.  The  third  side      ==         23-129 

43°  33'  44" 


The  other  angles    =    \    i  i°  ^4/ 


28" 


$  80.]  OBLIQJTE    TRIANGLES.  53 

Ratio  of  the  sum  of  the  two  sides  to  their  difference. 

5.  Given  two  sides  of  a  triangle  equal  to  77*245  and  92*341, 
and  the  angle  opposite  the  first  side  equal  to  124°  31'  48",  to 
solve  the  triangle. 

Ans.  The  question  is  impossible. 

6.  Given  two  sides  of  a  triangle  equal  to  75*486  and  92*341, 
and  the  angle  opposite  the  first  side  equal  to  55°  28'  12",  to 
solve  the  triangle. 

Ans.  The  question  is  impossible. 

80.  Theorem.  The  sum  of  two  sides  of  a  triangle 
is  to  their  difference,  as  the-  tangent  of  half  the  sum 
of  the  opposite  angles  is  to  the  tangent  of  half  their 
difference.   [B.  p.  13.] 

Proof.  We  have  (fig.  1.) 

a  :  b  =  sin.  A  :  sin.  B  ; 

whence,  by  the  theory  of  proportions, 

a  -j-  b  :  a  —  b  =  sin.  A  -f-  sin.  B  :  sin.  A  —  sin.  Bt 

which,  expressed  fractionally,  is 

a  -f-  b         sin.  A  ~(-  sin.  B 
a  —  b    '     sin.  A  —  sin.  B 

But,  by  (40), 

sin.  A  -f-  sin.  B       tang.  J  (A  -f-  B) 
sin.  A  —  sin.  B        tang.  %  (A  —  B)  ' 

whence 

a  +  b  __  tang.  J  (A  +  B) 
a  —  b       tang,  i  (A  —  B) 
5* 


(129) 


54  PLANE  TRIGONOMETRY.  [cH.  ft. 

Given  two  sides  and  the  included  angle. 

or 

a  +  b  :  a  —  6  =  tang.  J  (4  +  B)  :  tang.  £  (^4  —  B). 

81.  Problem.   To  solve  [a  triangle  when  two  of  its 
sides  and  the  included  angle  are  given.   [B.  p.  43.] 

Solution.  Let  the  two  sides  a  and  b  (fig.  1.)  be  given,  and 
the  included  angle  C,  to  solve  the  triangle. 

First.  To  find  the  other  two  angles.  Subtract  the  given 
angle  C  from  180°,  and  the  remainder  is  the  sum  of  A  and 
B,  for  the  sum  of  the  three  angles  of  a  triangle  is  180°,  that 
is, 

A  -f  B  =  180°  —  C, 
and 

£  (A  +  B)  —  90°  —  £  C  =  complement  of  J  C. 
The  difference  of  A  and  2?  is  then  found  by  (129) 

a  +  6:fl  —  b  =  tang.  J  (4  -f  jB)  :  tang.  %  (A  —  B). 

But  we  have 

tang.  J  (^4  -|-  B)  z=  cotan.  J  C; 
whence 

tang.  J  (.4— JB)  =  ^=|tang4(4+B)=^cotan.iG(180) 

The  greater  angle,  which  must  be  opposite  the  greater  side, 
is  then  found  by  adding  their  half  sum  to  their  half  difference  ; 
and  the  smaller  angle  by  subtracting  the  half  difference  from 
the  half  sum. 

Secondly.     The  third  side  is  found  by  the  proportion 
sin.  A  :  sin.  C  =z  a  :  c  ; 


§  82.]  OBLIQUE    TRIANGLES.  55 

Given  two  sides  and  the  included  angle. 

whence 

a  sin.  C 
sin.  A 


82.    Examples. 

1.  Given  two  sides  of  a  triangle  equal  to  99*341  and  1.234, 
and  their  included  angle  equal  to  169°  58',  to  solve  the  tri- 
angle. 

Solution.  Making  a  ==  99.341,  b  =  1.234 ;  and 

C  =z  169°  58',  J-  C  =  84°  59' ; 

we  have,  by  (130), 

a  +  b  —  100.575  (ar.  co.)  7.99751 

a  —  b  —    98.107  ^1.99170 

i  (A  +  B)  =   5°  1'  tang.    8.94340 


£  (A  —  B)  =  4°   53'  39"  tang.    8.93261 


^4=9°  54'  39" 

B  =z    0°  7/  21' 

a  =    99.341  1.99712 

C—  169°  58'  sin.  9.24110 

A  =  9°  54'  39"  cosec.  10.76416 


c  =  100.55  2.00238 

Ans.  The  third  side  =:  100.55 

T,       ..  ,  f    9°  54'  39' 

The  other  angles  z=  |    QO    ^  21„ 


56  PLANE  TRIGONOMETRY.  [CH.  VI. 

Segments  of  base  made  by  perpendicular  from  opposite  vertex. 

2.  Given  two  sides  of  a  triangle  equal  to  0.121  and  5.421 
and  the  included  angle  equal  to  1°  2' ;  to  solve  the  triangle. 

Ans.  The  other  side  =  5.294 

T,       ,.  .  i  178°  56'  35" 

lhe  other   angles  =  t       ~0     1/05// 

83.  Theorem.  One  side  of  a  triangle  is  to  the  sum 
of  the  other  two,  as  their  difference  is  to  the  difference 
of  the  segments  of  the  first  side  made  by  a  perpendicu- 
lar from  the  opposite  vertex,  if  the  perpendicular  fall 
within  the  triangle  ;  or  to  the  sum  of  the  distances  of 
the  extremities  of  the  base  from  the  foot  of  the  perpen- 
dicular, if  it  fall  without  the  triangle.   [B.  p.  14.] 

Proof  Let  AB  (figs.  12  and  13)  be  the  side  of  triangle 
ABC  on  which  the  perpendicular  is  let  fall,  and  BP  the  per- 
pendicular. 

From  5asa  centre  with  a  radius  equal  to  BC,  the  shorter 
of  the  other  two  sides,  describe  the  circumference  CC  E'E. 
Produce  AB  to  E1  and  ACto*C,  if  necessary. 

Then,  since  AC  and  AB  are  secants,  we  have, 

AC    AE  —  AE  :  AC. 


But 


and 


AE'  =  AB  +  BE  —AB  +  BC 

AE  =  AB  —  BE  =  AB  —  BC 

(fig.  12.)  AC  =  AP  —  PC  =  AP  —  PC 
(fig.  13.)  AC  =  AP  +  PC  —  AP  +  PC 


$  85.]  OBLIQ.UE    TRIANGLES.  57 

Given  the  three  sides. 

whence 

(fig.  12.)  AC :  AB  +  BC  =  AB  —  BC:  AP  —  PC 

(fig.  13.)  AC  :  AB  +  BC  =  AB  —  BC  :  AP  +  PC 

84.  Problem.    To    solve  a  triangle    when  its  three 
sides  are  given.  [B.  p.  43.] 

Solution.  On  the  side  b  (figs.  2  and  3.)  let  fall   the  perpen- 
dicular BP. 

Then,  by  §  83, 

(fig.  2.)  b  :  c  +  a  =  c  —  a  :  PA  —  PC 

(fig.  3.)  b  :  c  +  a  =  c  —  a  :  P^l  +  PC. 

These  proportions  give  the  difference  of  the  segments  (fig.  2.), 
or  their  sum  (fig.  3.).  Then,  adding  the  half  difference  to  the 
half  sum,  we  obtain  the  larger  segment  corresponding  to  the 
larger  of  the  two  sides  a  and  c.  And,  subtracting  the  half 
difference  from  the  half  sum, we  obtain  the  smaller  segment. 

Then,  in  triangles  BCP  and  ABP,  we  have,  by  (4)   and 
(92), 

A  AP 

cos.  A  =  ; 

c 

PC 

and  (fig.  2.)  cos.  C  s= , 

P  C 

(fig.  3.)  cos.  C  =  —  cos.  BCP  — . 

The  third  angle  B  is  found  by  subtracting  the  sum  of  A 
and  C  from  180°. 

85.  Corollary.  From  the   preceding  section,  we  have 


58  PLANE  TRIGONOMETRY.  [CH.  VI. 

Given  the  three  sides. 

(fig.  2.)  pa-pc=  («  +  «H«-g)  =  £lzi_«f 
(fig.  3.)  pa  +  pc  =  (c  +  ")(c-zA  =  £l^_«! 


which, 

added  to 

(% 

2.)  PA 

+  PC: 

-AC 

=  b 

(fig. 

3.)  PA 

—  PC: 

-AC: 

-  b 

gives 

2Pi  = 

c2  —  a* 

-  +  6  = 

b2  + 

c2  —  a2 

b 

b 

Hence 

PA=b2  +  c2- 
26 

and 


.  PA  &2  +  c2_a2 

cos.  A  =  —  =  -^_ (131) 


86.  Corollary.  If  (131)  is  cleared  from  fractions  it  becomes 
by  transposition 

a2  —  b2  +  c2  —  2  6c  cos.  ^4.  (132) 

87.  Corollary.  Add  unity  to  both  sides  of  (131)   and  we 
have 

1  +  cos.  it  *  ^mW-  +  1  -  267 

-  ¥b~c (133) 

Since  the  numerator  of   (133)   is  the    difference  of  two 
squares,  it  may  be  separated  into  two  factors,  and  we  have 


§  87.]  OBLIQ.UE    TRIANGLES.  59 

Given  the  three  sides. 

14.cosi_(H  c  +  a)(b  +  c-  a) 
1  +  cos.  A  _ 2^ • 

Now,  representing  half  the  sura  of  the  three  sides  of  a  tri- 
angle by  5,  we  have 

2sz=a  +  b  +  c,  (134) 

and 

2s-2a  =  2(5-«)=:fl  +  6  +c— 2  a—  b  +c—a.  (135) 

If  we  substitute  these  values  in   the  above  equation,  it  be- 
comes 

ii  a  4  s  (s  — a)        2s  (s  — a, 

1  +  cos.  A  =  — -  — .  (136) 

2  6c  be 

But,  by  (48), 

1  +  cos.  A  =  2  (cos.  %A)2. 
Hence 

or  (cos.J^)2=iiip^  (137) 

be 

cos.^=v(S(5~a))>  (138) 

which  corresponds  to  proposition  LXI.  of  B.  p.  14. 
In  the  same  way,  we  have 

cos.iB  =  */(tilzd)\  (139) 

coM<w(^).  (140) 


60  PLANE  TRIGONOMETRY.  [cH.  VI. 

Given  the  three  sides. 


88.  Corollary.  Subtract  both  sides  of  (131)  from  unity,  and 
we  have 

V*-\-&—a2  _  a2+2bc  —  b2  —  c2 


■  cos.  A  z=t  1  ■ 


26c  26c 

a2  —  (b— -c)2 
26~c"  ' 


(141) 


Since   the   numerator   of   (141)   is   the   difference  of  two 
squares,  it  may  be  separated  into  two  factors,  as  follows, 

{a — b  +  c)(a-\-b — c) 

1  —  cos.  A  =  * —Jt — ! -• 

26  c 

But  from  (134) 

25  —  2  6  =  2  (s  —  6)  =  a-J-6  +  c  —  2 6  =  a—  6  +  c  (142) 

2s  —  2c  =  2  (s  —  c)  =  a  +  b  +  c  —  2c  =  a+b  —  c.  (143) 

If  we  substitute  these  values  in  the  above  equations,  it  be- 
comes 

1  4(s  —  b)(s  —  c)        2(5  —  6)  (s  —  c)       t       . 

1  —  cos.  A  =  — - — f-- — '-  ==  — ^ ^-i '-.     (144) 

2  6c  6  c 

But,  by  (49), 

1  —  cos.  ^4  =  2  (sin.  \  A)2. 

Hence,  by  reduction, 

sin.M  =  v((S-6)6(;-^).  <145> 

In  the  same  way,  we  have 

si„.*B  =  v(^=4f-^)  (146) 

riMC=V(^a^'-*)).  ,(147) 


$  91.]  OBLIQUE    TRIANGLES.  61 


Given  the  three  sides. 


89.  Corollary.    The  quotients  of  (145,  146,  and  147),  di- 
vided by  (138,  139,  and  140),  are  by  (7) 

m»»b=v(('t(;'.1V)       <m 

90.  The  product  of  (136)  by  (144)  is 

i        /™«   A\2        ±s(s  —  a)(s  —  b)(s  —  c) 
1  —(cos.  A)2  =  p— . 

But  from  (10) 

1  —  (cos.  A)2  =  (sin.  A)2. 

Hence 

(sin.  A)2  =  4«(.-«)(.-t)(«-^ 


or 


1:^  W[s(s  —  a)(5  —  6)  (5  —  C)] 
b  c 


91.  Scholium.  The  problem  would  be  impossible,  if  the  given 
value  of  either  side  exceeded  the  sum  of  the  other  two. 


62  PLANE  TRIGONOMETRY.  [CH.   VI. 

Given  the  three  sides. 

92.    Examples. 

1.   Given   the  three  sides  of  a  triangle  equal  to  12.348, 
13.561,  14.091 ;  to  solve  the  triangle 

Solution*  First  Method. 

Make  (fig.  2.)  a  —  12.348  b  —  13.561 
c  L=  14.091. 
Then  by  §  84 

b  =  13.561  (ar.  co.)  8.86771 
c  -j-  a  =  26.439  1.42224 

c  —  a=    1.743  0.24130 


PA— PC  =  3.3982  0.53125 


PA  z=z  8.4796  0.92838 

PC —5.0814  0.70598 

c  =  14.091  (ar.  co.)8.85106 

a  =  12.348  (ar.  co.)    8.90840 


A  =  53°  0'  cos.  9.77944 


C  =  65°  42;  cos.  9.61438 

B  sa  180°  —  (A  +  C) 

=  180°  —  118°  42'  =  61°  18' 


*92-] 


OBLIQUE    TRIANGLES. 


63 


Given  the  three  sides. 


Second  Method. 
By  (138,  139,  and  140), 
a  =  12.348  (ar.  co.)  8.90840  (ar.  co.)  8.90840 

b  a=  13.561  (ar.  co.) 8.86771  (ar.  co.)  8.86771 

c  =  14.091  (ar.co.)8.85l06(ar.co.)8.85106  - 

1.30103 


s  =  20.000 
5-a=r7.652 
s-6=6.439 
5-cz=5.909 


COS. 


0.88377 


2  19.90357 


9.95179 


1.30103 


0.80882 


1.30103 


0.77151 


19.86931 


19.84865 


9.93466 


9.92433 


A  —  26°  30',  £  B  =  30°  39,  £  C  =  32°  51' 
4  =  53°  0',  JB  =  61°18/,   C=65°42/. 

.53°    0' 

Arts.     The  angles 


(53°    0' 

=  {  61°  18' 
I  65°  42. 


In  the  same  way  equations  (145-147)  would  furnish  a 
third  method,  (148-150)  a  fourth  method,  and  (151)  a  fifth 
method. 

2.  Given  the  three  sides  of  a  triangle  equal  to  17.856, 
13.349,  and  11.111  ;  to  solve  the  triangle. 

.  93°  19'  16" 

Ans.     The  angles 


(  93"  19'  lb" 

=  {  48°  16'  24" 
(  38°  24'  20". 


NAVIGATION    AND    SURVEYING. 


NAVIGATION    AND    SURVEYING. 


CHAPTER    I. 


PLANE    SAILING. 


1.  The  daily  revolution  of  the  earth  is  performed 
around  a  straight  line,  passing  through  its  centre,  which 
is  called  the  earth' s  axis. 

The  extremities  of  this  axis  on  the  surface  of  the 
earth  are  the  terrestrial  poles,  one  being  the  north  pole, 
and  the  other  the  south  pole. 

The  section  of  the  earth,  made  by  a  plane  passing 
through  its  centre  and  perpendicular  to  its  axis,  is  the 
terrestrial  equator,  [B.  p.  48.] 

2.  Parallels  of  latitude  are  the  circumferences  of 
small  circles,  the  planes  of  which  are  parallel  to  the 
equator. 

3.  Meridians  are  the  semicircumferences  of  great  cir- 
cles, which  pass  from  one  pole  to  the  other. 

The  first  meridian  is  one  arbitrarily  assumed,  to 
which  all  others  are  referred.     In  most  countries,  that 


68  NAVIGATION    AND    SURVEYING.  [cH.   I. 

Latitude.  Longitude. 

has  been  taken  as  the  first  meridian  which  passes 
through  the  capital  of  the  country.  But,  in  the  United 
States,  we  have  usually  adhered  to  the  English  custom, 
and  we  consider  the  meridian,  which  passes  through 
the  Observatory  of  Greenwich,  as  the  first  meridian. 
[B.  p.  48.] 

4.  The  latitude  of  a  place  is  its  angular  distance 
from  the  equator,  the  vertex  of  the  angle  being  at  the 
centre  of  the  earth  ;  or,  it  is  the  arc  of  the  meridian, 
passing  through  the  place,  which  is  comprehended  be- 
tween the  place  and  the  equator.  [B.  p.  48.] 

Latitude  is  reckoned  north  and  south  of  the  equator  from 
0°  to  90°. 

5.  The  difference  of  latitude  of  two  places  is  the 
angular  distance  between  the  parallels  of  latitude  in 
which  they  are  respectively  situated,  the  vertex  of  the 
angle  being  at  the  centre  of  the  earth  ;  or  it  is  the  arc 
of  a  meridian  which  is  comprehended  between  the  par- 
allels of  latitude.   [B.  p.  52.] 

The  difference  of  latitude  of  two  places  is  equal  to 
the  difference  of  their  latitudes,  if  they  are  on  the  same 
side  of  the  equator  ;  and  to  the  sum  of  their  latitudes, 
if  they  are  on  opposite  sides  of  the  equator.   [B.  p.  50.] 

6.  The  longitude  of  a  place  is  the  angle  made  by 
the  plane  of  the  first  meridian  with  the  plane  of  the 
meridian  passing  through  the  place ;  or  it  is  the  arc  of 
the  equator  comprehended  between  these  two  merid- 
ians. [B.  p.  48.] 


§   10.]  PLANE    SAILING.  69 

Distance.  Course.  Departure. 

Longitude  is  reckoned  East  and  West  of  the  first  meridian 
from0°  to  180°. 

7.  The  difference  of  longitude  of  two  places  is  the 
angle  contained  between  the  planes  of  the  meridians 
passing  through  the  two  places;  or  it  is  the  arc  of  the 
equator  comprehended  between  these  two  meridians. 

The  difference  of  longitude  of  two  places  is  equal 
to  the  difference  of  their  longitudes,  if  they  are  on  the 
same  side  of  the  first  meridian  ;  and  to  the  sum  of  their 
longitudes,  if  they  are  on  opposite  sides  of  the  first  me- 
ridian, unless  their  sum  be  greater  than  180°  ;  in  which 
case  the  sum  must  be  subtracted  from  360°  to  give  the 
difference  of  longitude,  [B.  p.  50. J 

8.  The  distance  between  two  places  in  Navigation 
is  the  portion  of  a  curve  which  would  be  described 
by  a  ship  sailing  from  one  place  to  the  other  in  a  path, 
which  crosses  every  meridian  at  the  same  angle. 
[B.  p.  52.] 

9.  The  course  of  the  ship,  or  the  bearing  of  the  twe 
places  from  each  other,  is  the  angle  which  the  ship's 
path  makes  with  the  meridian.  [B.  p.  52.] 

10.  The  departure  of  two  places  is  the  distance  o 
either  from  the  meridian  of  the  other,  when  they  are 
so  near  each  other  that  the  earth's  surface  may  be  con- 
sidered as  plane  and  its  curvature  neglected.     But,  if 
the  two  places  are  at  a  great  distance  from  each  other, 


70  NAVIGATION    AND    SURVEYING.  [CH.   I. 


Point.  Mariner's  compass. 

the  distance  is  to  be  divided  into  small  portions,  and 
the  departure  of  the  two  places  is  the  sum  of  the  de- 
partures corresponding  to  all  these  portions. 

11.  Instead  of  dividing  the  quadrant  into  90  degrees, 
navigators  are  in  the  habit  of  dividing  it  into  eight 
equal  parts  called  points;  and  of  subdividing  the  points 
into  halves  and  quarters.  A  point,  therefore,  is  equal 
to  one  eighth  of  90°,  or  to  11°  15'.   [B.  p.  52.] 

Names  are  given  to  the  directions  determined  by  the 
different  points, as  in  the  diagram  (fig. 14),  which  repre- 
sents the  face  of  the  card  of  the  Mariner's  Compass. 

The  Mariner's  Compass  consists  of  this  card,  at- 
tached to  a  magnetic  needle,  which  has  the  property 
of  constantly  pointing  toward  the  north  and  thereby 
determining  the  ship's  course. 

On  page  53  of  the  Navigator  a  table  is  given  of  the  angles 
which  every  point  of  the  compass  makes  with  the  meridian, 
and  on  page  169,  table  XXV.  the  log.,  sines,  &c.  are  given. 

12.  The  object  of  Plane  Sailing  is  to  calculate  the 
Distance,  Course  or  Bearing,  Difference  of  Latitude  and 
Departure,  when  either  two  of  them  are  known. 
[B.  p.  52.] 

13.  Problem.  To  find  the  difference  of  latitude  and 
departure^  when  the  distance  and  course  are  known. 
[B.  p.  54.] 

Solution.  First.  When  the  distance  is  so  small  that  the 
curvature  of  the  earth's  surface  may  be  neglected.     Let  A  B 


<§>   13.]  PLANE    SAILING.  71 

Given  distance  and  course. 

(fig.  15.)  be  the  distance.  Draw  through  A  the  meridian  AC, 
and  let  fall  on  it  the  perpendicular  BC.  The  angle  A  is  the 
course,  AC  is  the  difference  of  latitude,  and  BC  is  the  depart- 
ure.    Then,  by  (17,  and  18.) 

Diff.  of  lat.  =  dist.  X  cos.  course.  0^2) 

Departure  =  dist.  X   sin.  course.  (153) 

Secondly.  When  the  distance  is  great,  as  A  B  (fig.  16), 
then  divide  it  into  smaller  portions,  as  A  a,  a  b,  b  c,  &c. 
Through  the  points  of  division,  draw  the  meridians  AN,  an, 
bp,  &c.  Let  fall  the  perpendiculars  am,  b  n,  cp,  &c. 
Then,  as  the  course  is  every  where  the  same,  each  of  the 
angles  m  A  a,  n  a  b,  p  b  c,  &c.  is  equal  to  the  angle  A,  or  the 
course.  Moreover,  the  distances,  A  m,  a  n,  b  p,  &,c.  are  the 
differences  of  latitude  respectively  of  A  and  a,  a  and  b,  b  and 
c,  &c.  Also  a  m,  b  n,  c  p,  &c.  are  the  departures  of  the 
points  A  and  a,  a  and  b,  b  and  c,  &,c.  Therefore,  as  the 
difference  of  latitude  of  A  and  B  is  evidently  equal  to  the  sum 
of  these  partial  differences  of  latitude  ;  and  as  the  departure 
of  A  and  B  is  by  §  10  equal  to  the  sum  of  the  partial  depar- 
tures, we  have 

Diff.  of  lat.  =  Am-\-an-\-bp-\-  &c. 
Departure  =   am-\-bn-{-cp-\-  &c. 

But  the  right  triangles  m  A  a,  n  ab,p  b  c,  &c.  give   by  (152, 
and  153.) 

A  m  —  A  a  X  cos.  course,  a  m  =  A  a  X  sin.  course  ; 
a  n  =  a  b  X  cos.  course,  b  n  =  a  b  X  sin.  course  ; 
b p  =  b  c  X  cos.    course,  cp  =    b  c  X  sin.  course. 
&c.  &c. 


72  NAVIGATION    AND    SURVEYING.  [CH.   I. 

Given  course  and  departure. 

The  sums  of  these  equations  give 

Diff.  of  lat.  =  i?a-)-«n  +  &c  +  &c 

=  (Aa-\-ab-\-bc-\-  &c.)  X  cos.  course, 
Departure    =  a  m  -\-  b  n  -\-  c  p  ~\-  &c. 

=  (Aa-\-ab-{-bc-\-  &c.)  X    sin.  course. 
But 

Aa-\-ab-\-bc-\-  &c.  =  AB  =  distance. 
Hence, 

Diff.  of  lat.  =  dist.  X  cos.  course, 
Departure    =  dist.  X  sin.  course ; 

precisely  the  same  with  (152)  and   (153). 

This  shows  that  the  method  of  calculating  the  dif- 
ference of  latitude  and  departure  is  the  same  for  all 
distances,  and  that  all  the  problems  of  Plane  Sailing 
may  be  solved  by  the  right  triangle  (fig.  15.)  [B.  p.  52.] 

A  table  of  difference  and  latitude  and  departure  are  given  in 
pages  1-6,  Tables  I.  and  II.  of  the  Navigator,  which  might 
be  calculated  by  (152  and  153.) 

14.  Problem.  To  find  the  distance  and  difference  of 
latitude,  when  the  course  and  departure  are  known. 
[B.  p.  55.] 

Solution.  There  are  given  (fig.  15.)  the  angle  A  and  the 
side  B  C.     Hence,  by  (19,  and  20), 

Distance  ==  departure  X  cosec.  course.  (154) 

Diff.  of  lat.  =  departure  X  cotan.  course.  (*55) 


§   18.]  PLANE    SAILING.  73 

Cases  of  plane  sailing. 

15.  Problem.  To  find  the  distance  and  departure, 
when  the  course  and  difference  of  latitude  are  known. 
[B.  p.  55.] 

Solution.  There  are  given  (fig.  15.)  the  angle  A  and  the 
side  AC.     Then,  by  (21,  and  22). 

Distance  =  diff.  of  lat.  X  sec-  course.  (156*) 

Departure  =  diff.  of  lat.  X  tang,  course.  (157) 

16.  Problem.  To  find  the  course  and  difference  of 
latitude,  when  the  distance  and  departure  are  known. 
[B.  p.  57.] 

Solution.  There  are  given  (fig.  15.)  the  hypothenuse  AB 
and  the  side  EC.  Then,  by  (23,  and  25), 

departure 

sin.  course  z=  — -? ,  (15o) 

distance 

Diff.  of  lat.  z=  \/  [(dist.2)  —  (departure)2].         (159) 

17.  Problem.  To  find  the  course  and  departure,  ivhen 
the  distance  and  difference  of  latitude  are  known. 
[B.  p.  56.] 

Solution.  There  are  given  (fig.  15.)  the  hypothenuse  AB 
and  the  leg  AC.     Then,  by  (23  and  25), 

diff  of  lat. 

cos.  course  =  — ,  (160) 

distance.  ' 

Departure  =  s/  [(dist.)2  _  (diff.  of  lat.)2].        (161) 

18.  Problem.  To  find  the  course  and  distance,  when 
the  departure  and  difference  of  latitude  are  known. 
[B.  p.  57.] 

7 


74  NAVIGATION    AND    SURVEYING.  [CH.   I. 

Examples. 

Solution.  There  are  given  (fig.  15.)  the  legs  AC  and  BC. 
Then, 

departure  ,-tLL* 

tang,  course  ^5I^___  (162) 

Dist.  =  diff.  of  lat.  X  sec.  course.  (163) 

19.    Examples. 

1.  A  ship  sails  from  latitude  3°  45'  S.,  upon  a  course 
N.  by  E.,  a  distance  of  2345  miles ;  to  find  the  latitude  at 
which  it  arrives,  and  the  departure  which  it  makes. 

Ans.     Latitude       —       34°  35'  N. 

Departure    =z       458  miles. 

2.  A  ship  sails  from  latitude  62°  19'  N.,  upon  a  course 
W.  N.  W.,  till  it  makes  a  departure  of  1000  miles ;  to  find  the 
latitude  at  which  it  arrives,  and  the  distance  sailed. 

Ans.     Latitude       ==       69°  13'  N. 

Distance      =       1082  miles. 

3.  The  bearing  of  Paris  from  Athens  is  N.  54°  56'  W. ; 
find  the  distance  and  departure  of  these  two  places  from  each 
other. 

Ans.     Distance       =       1135  miles. 
Departure     =       929  miles. 

4.  A  ship  sails  from  latitude  72°  3'  S.  a  distance  of 
2000  miles,  upon  a  course  between  the  north  and  the  west, 
that  is,  northwesterly,  until  it  makes  a  departure  of  1000 
miles ;  find  the  latitude  at  which  it  arrives,  and  the  course. 


§  19.]  PLANE    SAILING.  75 

Examples. 

Ans.     Latitude     =     43°  IT  S. 
Course       =     N.  30°  W. 

5.  The  distance  from  New  Orleans  to  Portland  is  958  miles; 
find  the  bearing  and  departure. 

Ans.     Bearing       =     N.  49°  24'  E. 

Departure  as     1257  miles. 

6.  The  departure  of  Boston  from  Canton  is  8790  miles; 
find  the  bearing  and  distance. 

Ans.     Bearing       as     N.  82°  31'  E. 

Distance     =     8865  miles. 


76  NAVIGATION    AND    SURVEYING.  [CH.   II. 

Traverse. 


CHAPTER   II. 


TRAVERSE    SAILING. 


20.  A  traverse  is  an  irregular  track  made  by  a  ship 
when  sailing  on  several  different  courses. 

The  object  of  Traverse  Sailing  is  to  reduce  a  trav- 
erse to  a  single  course,  where  the  distances  sailed  are  so 
small  that  the  earth's  surface  may  be  considered  as  a 
plane.   [B.  p.  59.] 

21.  Problem.  To  reduce  several  successive  tracks  of 
a  ship  to  one  ;  that  is,  to  find  the  single  track,  leading 
to  the  place,  which  the  ship  leas  actually  reached,  by 
sailing  on  a  traverse.   [B.  p.  59.] 

Solution.  Suppose  the  ship,  to  start  from  the  point  A  (fig. 
17.)  and  to  sail,  first  from  A  to  B,  then  from  B  to  C,  then 
from  C  to  JEJ,  and  lastly  from  E  to  F ;  to  find  the  bearing  and 
distance  of  F  from  A.  Call  the  differences  of  latitude,  cor- 
responding to  the  1st,  2d,  3d,  and  4th  tracks,  the  1st,  2d,  3d, 
and  4th  differences  of  latitude ;  and  call  the  corresponding 
departures  the  1st,  2d,  3d,  and  4th  departures.  Then  we 
need  no  demonstration  to  prove  that, 

Diff.  of  lat.  of  A   and  P=  1st  diff.  of  lat.  —  2d  diff.  of  lat. 
+  3d  diff.  of  lat.  —  4th  diff.  &c. ; 

or  that  the  difference  of  latitude  of  A  and  F  is  found 


§  22.]  TRAVERSE    SAILING.  77 

To  reduce  a  traverse  to  a  single  course. 

by  taking  the  sum  of  the  differences  of  latitude  corre- 
sponding to  the  northerly  courses,  and  also  the  sum  of 
those  corresponding  to  the  southerly  courses,  and  the 
difference  of  these  sums  is  the  required  difference  of 
latitude. 

By  neglecting  the  earth's  curvature,  we  also  have, 

Dep.  of  A  and  F  =  1st  dep.  —  2d  dep.  —3d  dep.  +  4th  dep. 

or  the  departure  of  A  and  F  is  found  by  taking  the 
sum  of  the  departures  corresponding  to  the  easterly 
courses,  also  the  sum  of  those  corresponding  to  the  west- 
erly courses  ;  and  the  difference  of  these  sums  is  the 
required  departure. 

Having  thus  found  the  difference  of  latitude  and  de- 
parture of  A  and  F,  their  distance  and  bearing  are 
found  by  <§>  18. 

22.  The  calculations  of  traverse  sailing  are  usually 
put  into  a  tabular  form,  as  in  the  following  example. 
In  the  first  column  of  the  table  are  the  numbers  of 
the  courses  ;  in  the  second  and  third  columns  are  the 
courses  and  distances  ;  in  the  fourth  and  fifth  columns 
are  the  differences  of  latitude,  the  column,  headed  N, 
corresponding  to  the  northerly  courses,  and  that  headed 
S,  to  the  southerly  courses;  in  the  sixth  and  seventh 
columns  are  the  departures,  the  column,  headed  E,  cor- 
responding to  the  easterly  courses,  and  that,  headed  W; 
to  the  westerly  courses.  [B.  p.  59.] 

7* 


78 


NAVIGATION    AND    SURVEYING. 


[CH.  11. 


Examples. 


23.    Examples. 

1.  A  ship  sails  on  several  successive  tracks,  in  the  order 
and  with  the  courses  and  distances  of  the  first  three  columns 
of  the  following  table ;  find  the  bearing  and  distance  of  the 
place  at  which  the  ship  arrives,  from  that  from  which  it 
started. 


No. 

Course. 

Dist. 

N. 

S. 

E. 

W. 

1 

N.  N.  E. 

30 

27.7 

115 

2 

N.  W. 

80 

56.6 

56.6 

3 

West. 

60 

60.0 

4 

S.  E.  by  S. 

55 

45.7 

30.6 

5 

North. 

43 

43.0 

6 

S.  by  W. 

152 

149.1 
194.8 

42  1 

29.7 

Sum  of  col 

umns, 

127.3 

146.3 

127.3                 42.1 



Diff.  lat.  =  67.5S.dep.=104.2W. 
Dep.  =  104.2  2.01787 

Diff.  of  lat  =    67.5  (ar.  co.)  8.17070  1.82930 


Bearing  =  57°  4'    tang.      0.18857  sec. 


0.26467 


Dist.  =  124  1  2.09397 

Ans.       Bearing  =  S.  57°  4'  W. 
*  Distance  =  124.1  miles. 

2.  A  ship  sails  on  the  following  successive  tracks,  South  10 
miles,  W.  S.  W.  25  miles,  S.  W.  30  miles,  and  West  20  miles  ; 
it  is  bound  to  a  port  which  is  at  a  distance  of  100  miles  from 
the  place  of  starting,  and  its  bearing  is  W.  by  N. 


§  23,]  PLANE    SAILING.  79 

Examples. 

Required  the  bearing  and  distance  of  the  port  to  which  the 
ship  is  bound,  from  the  place  at  which  it  has  arrived. 

Ans.     Bearing  =5  S.  51°  47'  W. 
Distance  =  239  miles. 


80  NAVIGATION    AND    SURVEYING.  [CH.  III. 

Differences  of  longitude  in  parallel  sailing. 


CHAPTER    III. 

PARALLEL    SAILING. 

24.  Parallel  Sailing  considers  only  the  case  where 
the  ship  sails  exactly  east  or  west,  and  therefore  re- 
mains constantly  on  the  same  parallel  of  latitude.  Its 
object  is  to  find  the  change  in  longitude  corresponding 
to  the  ship's  track.   [B.  p.  63.] 

25.  Problem.  To  find  the  difference  of  longitude  in 
parallel  sailing.  [B.  p.  65.] 

Solution.  Let  AB  (fig.  18.)  be  the  distance  sailed  by  the 
ship  on  the  parallel  of  latitude  A  B.  As  the  course  is  exactly 
east  or  west,  the  distance  sailed  must  be  itself  equal  to  its  de- 
parture. 

The  latitude  of  the  parallel  is  AD  A'  or  A  A1.  The  angle 
AEB  =  A'D  B1,  or  the  arc  A'  B',  is  the  difference  of  lon- 
gitude. Denote  the  radius  of  the  earth  A'D  =  B'D  =  A  D 
by  JR,  and  the  radius  of  the  parallel  AE  =  BE  by  r ;  also 
the  circumference  of  the  earth  by  C,  and  that  of  the  parallel 
by  c. 

Since  AB  and  A'B'  correspond  to  the  equal  angles  AE  B 
and  A'D  B\  they  must  be  similar  arcs,  and  give  the  propor- 
tion, 

AB  :AB'  =  c  :  C, 
or  Dep.  :  diff.  long.  =  c  :  G 


<§>  27.]  PARALLEL    SAILING.  81 

Difference  of  longitude. 

But,  as  circumferences  are  proportional  to  their  radii, 
c  :  C=r  :R. 
Hence,  leaving  out  the  common  ratio, 

Dep.  :  cliff,  long.  =  r  :  JR. 

Putting  the  product  of  the  extremes  equal  to  that  of  the 
means, 

r.  diff.  of  long.  =  R.  departure. 

But,  in  the  triangle  AD E,  since 

DAE=ADA=  latitude, 

we  have,  from  (17), 

r  =  R  X  cos.  lat. 

which,  substituted  in  the  above  equation,  gives,  if  the  result  is 
divided  by  R, 

Diff.  long.  X  cos.  lat.  =  departure.  (1°*4) 

Hence,  by  (8), 

Diff.  long,  =  ^R^™  =  dep.  X  sec.  lat.         (165) 
cos.  lat.  r  x       ' 

26.  Corollary.  Since  the  distance  is  the  same  as  the  depar- 
ture in  parallel  sailing.  The  word  distance  may  be  substituted 
for  departure  in  (164)  and  (165). 

27.  Corollary.  It  appears,  from  (164)  and  (165),  that 
if  a  right  triangle  (fig.  18.)  is  constructed,  the  hypothe- 
nuse  of  which  is  the  difference  of  longitude,  and  one  of 
the  acute  angles  the  latitude,  the  leg  adjacent  to  this 
angle  is  the  departure.  All  the  cases  of  parallel  sailing 
may ,  then,  be  reduced  to  the  solution  of  this  triangle. 


82  NAVIGATION    AND    SURVEYING.  [CH.   III. 

Differences  of  places  on  the  same  parallel. 

28.  Problem.  To  find  the  distance  between  two  places 
which  are  upon  the  same  parallel  of  latitude. 

Solution.  This  problem  is  solved  at  once  by  (164). 

29.  The  Table,  p.  64,  of  the  Navigator,  which  "  shows  for 
every  degree  of  latitude  how  many  miles  distant  two  meridians 
are,  whose  difference  of  longitude  is  one  degree,"  is  readily 
calculated  by  this  problem. 


30.    Examples. 

1.  A  ship  sails  from  Boston   1000  miles  exactly  east ;  find 
the  longitude  at  which  it  arrives. 

Ans.     Longitude  sought  =  51°  48'  W. 

2.  Find  the  distance  of  Barcelona  (Spain)  from  Nantucket 
(Massachusetts). 

Ans.     Distance     =     3252  miles. 

3.  Find  the  distance  between  two  meridians,  whose  differ- 
ence of  longitude  is  one  degree  in  the  latitude  of  45°. 

Ans.     Distance     =     42.43  miles. 


§  32.]  MIDDLE    LATITUDE    SAILING.  83 

Middle  latitude. 


CHAPTER   IV. 


MIDDLE    LATITUDE    SAILING. 


31.  The  object  of  Middle  Latitude  Sailing  is  to  give 
an  approximative  method  of  calculating  the  difference 
of  longitude,  when  the  difference  of  latitude  is  small. 
[B.  p.  66.] 

32.  Problem.  To  find  the  difference  of  longitude  by 
Middle  Latitude  Sailing,  when  the  distance  and  course 
are  known,  and  also  the  latitude  of  either  extremity  of 
the  ship's  track.   [B.  p.  71.] 

Solution.  The  difference  of  latitude  and  departure  are  found 
by  (152)  and  (153), 

Diff.  lat.  =  dist.  X  cos.  course 

Departure  =  dist.  X  sin.  course. 

The  difference  of  longitude  may  then  be  found  by  means  of 
(165).  But  there  is  a  difficulty  with  regard  to  the  latitude  to 
be  used  in  (165);  for,  of  the  two  extremities  of  the  ship's 
track,  the  latitude  of  one  is  smaller,  while  the  latitude  of  the 
other  extremity  is  larger  than  the  latitude  of  the  rest  of  the 
track.  Navigators  have  evaded  this  difficulty  by  using  the 
Middle  Latitude  between  the  two,  as  sufficiently  accurate, 
when  the  difference  of  latitude  is  small.  Now  the  middle  lat- 
itude is  the  arithmetical  mean  between  the  latitudes  of  the 
extremities,  so  that  we  have, 


84  NAVIGATION    AND    SURVEYING.  [CH.  IV. 

To  find  the  difference  of  longitude. 

Middle  lat.  =  J  sum  of  the  lats.  of  the  extremities  of  the 
track ;  (166) 

and,  by  (165), 

departure 

DifF.  long,  b  E-— —  ==  dep.  X  sec.  mid.  lat.    (167) 

&         cos.  mid.  lat.  ' 

or,  by  substituting  (153), 

DifF.  long.  =  dist.  X  sin.  course  X  sec.  mid.  lat.      (168) 

"  This  method  of  calculating  the  difference  of  longitude  may 
be  rendered  perfectly  accurate  by  applying  to  the  middle  lat- 
itude a  correction,"  which  is  given  in  the  Navigator,  and  the 
method  of  computing,  which  will  be  explained  in  the  suc- 
ceeding chapter.     [B.  p.  76,] 

33.  By  combining  the  triangle  (fig.  16.)  of  Plane 
sailing  with  that  (fig.  18.)  of  Parallel  sailing  a  triangle 
(fig.  19.)  is  obtained,  by  which  all  the  cases  of  Middle 
Latitude  sailing  may  be  solved. 

34.  Problem.  To  find  the  distance  and  bearing  of 
two  places  from  each  other,  when  their  latitudes  and 
longitudes  are  known.    [B.  p.  68.] 

Solution.    From  (fig.  19)  we  have 

Departure  =  diff.  long.  X  cos.  mid.  lat.  (169) 

.       .  .  departure 

tang,  bearing  =  ^^^-  (170) 

dist.  ==  diff.  lat.  X  sec.  bearing.  (J 71) 


§  38.]  MIDDLE    LATITUDE    SAILING.  85 

Cases  of  middle  latitude  sailing. 

35.  Problem.  To  find  the  course,  distance,  and  dif- 
ference of  longitude,  when  both  latitudes  and  the  depar- 
ture are  given.    [B.  p.  70.] 

Solution.  The  difference  of  longitude  is  found  by  (167),  the 
course  by  (170),  and  the  distance  by  (171). 

36.  Problem.  To  find  the  departure,  distance,  and 
difference  of  longitude,  when  both  latitudes  and  the 
course  are  given,   [B.  p.  72.] 

Solution.  The  departure  is  found  by  the  formula 

departure  —  diff.  lat.  X  tang,  course;  (172) 

the  distance  by  (171);  and  the  difference  of  longitude  may 
be  found  by  (167),  or  by  substituting  (172)  in  (167) 

diff.  long.  z=z  diff.  lat.  X  tang,  course  X  sec.  mid.  lat.    (173) 

37.  Problem.  To  find  the  course,  departure,  and  dif- 
ference of  longitude,  when  both  latitudes  and  the  distance 
are  given.    [B.  p.  73.] 

Solution.    The  course  is  found  by  the  formula 

diff.  lat.  /t_ 

cos.  course  ==  — ;  (174) 

dist.  v       ' 

the  departure  by 

departure  zsz  dist.  X  sin.  course;  (175) 

and  the  difference  of  longitude  by  (167). 

38.  Problem.   To  find  the  difference  of  latitude,  dis- 

8 


86  NAVIGATION    AND    SURVEYING.  [CH.  IV. 

Examples. 

tance,  and  difference  of  longitude,  when  one  latitude, 
course,  and  departure  are  given.    [B.  p.  74.] 

Solution.  The  difference  of  latitude  is  found  by  the  formula 

diff.  lat,  =  dep.  X  cotan.  course;  (176) 

the  distance  by  the  formula 

dist.  =  dep.  X  cosec.  course;  (177) 

and  the  difference  of  longitude  by  (167). 

39.  Problem.  To  find  the  course,  difference  of  lati- 
tude, and  difference  of  longitude,  when  one  latitude,  the 
distance,  and  departure  are  given.']    B.  p.  75.] 

Solution.  The  course  is  found  by  the  formula 

sin.  course  ±z    ,.      ;  (178) 

dist.  v 

the  difference  of  latitude  by  the  formula 

diff.  lat.  =  dist.  X  cos.  course;  (179) 

and  the  difference  of  longitude  by  (167). 


40.    Examples. 

1.  A  ship  sailed  from  Halifax  (Nova  Scotia)  a  distance  of 
2509  miles,  upon  a  course  S.  79°  34'  E. ;  find  the  place  at 
which  it  arrived. 

Solution.    By  §  32, 


<§>  40.]                      MIDDLE    LATITUDE    SAILING.  87 

Examples. 

dist.  =  2509                            3-39950  3.39950 

course  ==  79°  34'                 cos.  9.25790  sin.  9.99276 


diff.  lat.  =  454'  =  7°   34'  S.  2.65740 

given  lat.  —  44°  36'  N.  mid.  lat.  =  40°  49' 

required  lat.  =37°   2'N.      cor.     —         7' 


cor.  mid.  lat.     =40°  56'  sin.  10.12178 


diff.  long.  ==  3266'  —  54°  26'  E.  3.51404 

given  long.  =  63°  28'  W. 

required  long.  tk    9°    2'  W. 

Ans.     The  place  arrived  at   is  one  mile  south  of  Cape  St. 
Vincent  in  Portugal. 

2.  Find  the  bearing  and  distance  of  Canton  from  Washing- 
ton. 

Solution.    By  §  34, 
lat.ofWashington^SS^'N.  long.  =    77°    3  W. 

lat.  of  Canton         =23°   7'N.  long.  =:  113°  14' E. 


diff.  lat.  =  946'  —  15°  46,  sum  of  longs.  =  190°  17' 
mid.  lat.  =  31°   0'     diff.  long.      =  169°  43=10183/ 

cor.  =  31' 


cor.  mid.  lat.  =  31°  31'  cos.    9.93069 

diff.  long.  =  10183'  4.00788 

diff.  lat.     =      946'  ar.co.    7.02411  2.97589 


bearing     =  S.  83°  47' W.         tang.  10.96268   sec.  10.96526 
dist.     —  8733  miles  3.94115 


NAVIGATION    AND    SURVEYING.  [CH.  1V« 


Examples. 


3.  A  ship  sails  from  New  York  a  distance  of  675J  miles, 
upon  a  course  S.  E.  £  S.  ;  find  the  place  at  which  it  arrives. 

Ans.  Three  miles  to  the  west  of  Georgetown  in  Bermuda. 

4.  Find  the  bearing  and  distance  of  Portland  (Maine)  from 
New  Orleans. 

Ans.         The  bearing     —     N.  49°  25'  E. 

The  distance    sc     1257  miles. 

5.  A  ship  from  the  Cape  of  Good  Hope  sails  northwesterly 
until  its  latitude  is  22°  3' S.,  and  its  departure  3110  miles  I 
find  its  course,  distance  sailed,  longitude,  and  its  distance  from 
Cape  St.  Thomas  (Brazil). 

Ans.  Course         =  N.  76°  38'  W. 

Distance      —  3197  miles. 
Longitude    r=  18°  W. 
Distance  to  the  Cape  St.  Thomas  z=  22  miles. 

6.  A  ship  sails  from  Boston  upon  a  course  E.  by  N.  until  it 
arrives  in  latitude  45°  21'  N. ;  find  the  distance,  its  longitude, 
and  its  distance  and  bearing  from  Liverpool. 

Ans.         Distance  sailed  =  920  miles. 

Longitude  —  50°  10'  W. 

Distance  from  Liverpool  =  1905  miles 

Bearing  from  Liverpool  =:  S.  75°  22'  W. 

7.  A  ship  sails  southwesterly  from  Gibraltar  a  distance  of 
1500  miles,  when  it  is  in  latitude  14°  44'  N. ;  find  its  course 
and  longitude  and  distance  from  Cape  Verde. 


$  40.]  MIDDLE    LATITUDE    SAILING.  89 

Examples. 

Ans.         Course         =  S.  37°  21'  W. 

Longitude   =       18°    3' W. 

Dist.  from  Cape  Verde  =r       339  miles. 

8.  A  ship  sails  from  Nantucket  upon  a  course  S.  62°  ll7  E., 
until  its  departure  is  2274  miles  ;  find  the  distance  sailed,  and 
the  place  arrived  at. 

Ans.         Distance     s=     2571  miles. 

The  place  arrived  at  is  261  miles  north  of  Santa  Cruz. 

9.  A  ship  sails  southwesterly  from  Land's  End  (England),  a 
distance  of  3461  miles,  when  its  departure  is  3300  miles  ;  find 
the  course  and  the  place  arrived  at. 

Ans.         The  course    =     S.  72°  27'  W. 

The  place  arrived  at  is  Charleston  (South  Carolina) 
Light  House. 


8* 


90  NAVIGATION    AND    SURVEYING.  [CH.  V. 

To  find  the  difference  of  longitude. 


CHAPTER    V. 


MERCATOR  S    SAILING. 


41.  The  object  of  M creator's  Sailing  is  to  give  an 
accurate  method  of  calculating  the  difference  of  longi- 
tude.   [B.  p.  78.] 

42.  Problem.  To  find  the  difference  of  longitude,  when 
the  distance,  the  course,  and  one  latitude  are  known. 

Solution.  Let  A  B  (fig.  16)  be  the  ship's  track.  Divide  it 
into  the  small  portions  A  a,  ab,  be,  &c,  which  are  such  that 
the  difference  of  longitude  is  the  same  for  each  of  them,  and 
let 

d  —  this  small  difference  in  longitude. 

Let  also 

L  —  the  latitude  of  A, 
L1—  the  latitude  of  B, 

I  —  the  latitude  of  one  of  the  points  of  division  as  b, 

I'  =  the  latitude  of  the  next  point  c, 

C  =  the  course. 

The  distance  b  c  may  then  be  supposed  so  small,  that  the  for- 
mulas of  middle  latitude  sailing  may  be  applied  to  it ;  and 
(173)  gives 


§  42.]  mercator's  sailing.  91 

Difference  of  longitude. 

dz=(l'—l)X  tang.  C  X  sec.  J  (/'  +  l)t  (180) 

or 

i  d  cotan.  C=  — ^ir~?A   '  <181) 

cos,  J  (/  +  0 

If,  now,  the  mile  is  adopted  as  the  unit  of  length,  and  if 

R  =  the  earth's  radius  in  miles,  (182) 

J7~    —  is  the  leugth  of  the  arc  J  (V  —  Z), 

expressed  in  terms  of  the   radius  as  unity ;  and  since  this  arc 
is  very  small,  its  length  is  by  §  22  equal  to  its  sine  or 

i{l'~  l)  =  sin.  i(l'-l);  ( 183) 


which  substituted  in  (181)  gives 

d  cotan.  C  _      sin.  £  {I1  —  Z) 


(184) 


(185) 


2JR  cos.  \  (/'  +  *) 

Let  now 

d  cotan.  C  sin.  J  (Z'  —  Z ) 

~~2BT       =  co¥.  £(/'  +  Z) ; 

and  (185)  may  be  written  in  the  usual  form  of  a  proportion 

sin.  i  (I'  —  Z)  :  cos.  £  (Z'  +  Z)  =  m  :  1 ;  (186) 

whence,  by  the  theory  of  proportions 


cos,  j  (V  +  I)  +  sin.  £  (/'  —  /)  __   1  +  i» 
cos.  £  (/'  +  Z)  —  sin.  i  (Z'  —  I)  ~~   \—m 


(187) 


But  if  in  (40)  we  put 

4  =  90°-  l(l'  +  l),B  =  £  (I'  — I);  (188) 

we  have 


92  NAVIGATION    AND    SURVEYING.  [CH.  V. 

Difference  of  longitude. 

A  +  B  =  90°  —  I,  A  —  B  =  90°  —  /',  (189) 

and  (46)  becomes 

cos,  i  (f  +  /)  +  bid,  j  (/'  —  /)  =  cotan.  (45°  — jf)  . 
cos.  i  (/'  +  /)  —  sin.  J  (/'  —  I)      cotan.(45°  —  £1)'    K       ' 
whence,  if  we  put 

ar=I±A  (i9i) 

1  —  m 
colzn.JW.-in   =M  (1W) 


cotan.  (45°  —  i  / ) 

Hence,   the  successive    values    of  cotan.  (45°  —  £  I)    at   the 
points  Ay  a,  6,  &c,  form  a  geometric  progression;  and  if 

D  =  the  difference  of  longitude  of  A  and  B, 
n  =  the  number  of  portions  of  AB ; 
we  have  by  (185) 

n  a.  5.  ==*  g  s    ^ -,  (193) 

d        2  Rm  tang.  C  v       ' 

and  by  the  theory  of  geometric  progression 

cotan.  (45°  —  £  L1)  ==  cotan.  (45°  —  £  L)  W,      (194) 

and  by  logarithms 

log.  cotan.  (45°— ££.')— loS- cotan-  (45°  —  lL)=\°g-  Mn.  ( 195) 
If,  lastly,  we  put 

e  =  M&%  (196) 

we  have 


Mn  s*e**H&  (197) 


§  43.]  mercator's  sailing.  •  93 

Meridional  difference  of  latitude. 

which  substituted  in  (195)  gives  by  a  simple  reduction 

\j£i  1°g-cotan-(45°-i^/)-1^  log.cotan.(45°-jZ)] 
X  tang.  C  =  D  (199) 

Now  the  value  of— «—  log.  cotan.  (45°  — J  L)  has 

been  calculated  for  every  mile  of  latitude,  and  inserted 
in  tables.  [B.  Table  III.]  It  is  called  the  Meridional 
Parts  of  the  Latitude,  and  the  method  of  computing  it- 
is  given  in  the  following  section. 

The  difference  between  the  meridional  parts  of  the  two 
latitudes,  when  the  latitudes  are  both  north  or  both 
south,  is  called  the  Meridional  Difference  of  Latitude  ; 
but  when  one  of  the  latitudes  is  north  and  the  other 
south,  the  sum  of  the  meridional  parts  is  the  meridional 
difference  of  latitude. 

Hence  (199)  gives 

D  z=  diff.  long.  =  mer.  diff.  lat.  X  tang  course.        (200) 

43.  Corollary.  The  difference  of  longitude  is  as  in 
(fig.  20.)  the  leg  DE  of  a  right  triangle,  of  which  AD 
is  the  meridional  difference  of  latitude,  and  the  angle  A 
the  course  ;  and  by  combining  this  triangle  with  the 
triangle  ABC  of  plane  sailing,  all  the  cases  of  Mer ca- 
tor's  Sailing  are  reduced  to  the  solution  of  these  two 
similar  right  triangles. 


94  •  NAVIGATION    AND    SURVEYING.  [CH.   V. 

Table  of  meridional  parts. 

44.  Problem.  To  calculate  the  Table  of  Meridional 
Parts. 

Solution.  I.  The  value  of  R  is  found  from  the  considera- 
tion that  if 

n  =  the  ratio  of  a  circumference  to  its  diameter 
=  3.1416,  (201) 

we  have, 

2  TV  R  —  the  circumference  of  the  earth 
=  360°  =  21600' 
and 

R  =  3^-  m  3437.7.  (202) 

II.  In  finding  the  value  of  e,  the  portions  of  the  distance 
are  supposed  to  be  infinitely  small,  hence  m  is  by  (185)  also 
infinitely  small,  and  its  reciprocal  is  infinitely  great. 

If  I  -\-  mis  divided  by  1  —  m,  as  follows, 

l_w)l4-m(l+2m  +  2m2  +  &c. 
1  —  m 


+  2m 

2  m  —  2  m2 

+  2m2 

2  m2  —2  m3 

+  2m3 
we  have  by  (191) 

M=  1  +  2  m  +  2  m2  +  &c.  (203) 

But  since  m  is  infinitely  small,  m2,  m3,  &c.   are  infinitely 


M4.] 


MERCATOR  S    SAILING. 


95 


Table  of  meridional  parts. 


smaller,  and  the  error  of  rejecting  them  in  (203)  is  less  than 
any  assignable  quantity  ;  whence 


if/=l+2m 

and  by  the  binomial  theorem,  and  (196), 

i 
e  =  (1  +2ffl)s 


(204) 


(2  m)2 


2  m  \  2  m  /  \  2  m  /  1 .  2  .  3     '  v       ' 


1 

But  -— 

2  hi 

ble  error, 

1 

2m 


is  infinite  and  gives  therefore,  without  any  assigna- 
l=-^-,J-     -2=  A-.&c-         (206) 


2  m '   2  m 
which,  substituted  in  (205),  gives 


6=1 


2m 


.2> 


I 


(2m)s 


(2  m)  s 


+  &c. 


"(2m)2     1.2      '    (2m)3  1.2.3 

+  &c.  (207) 


=  i  +  i  +  _i_  j L 1 1 

^     ^1.2^1.2.3^1.2.3.4 


so  that  e  is  the  sum  of  a  series  of  terms,  the  first  of 
which  is  unity )  and  each  succeeding  term  is  obtained 
by  dividing  the  'preceding  term  by  the  place  of  this  pre- 
ceding term. 


96  NAVIGATION    AND    SURVEYING.  [CH. 

Neperian  logarithms.  Table  of  meridional  parts. 


The  value  of  e  is  thus 

computed. 

i) 

1  .  000000 

2) 

1  .  000000 

3) 

.  500000 

4) 

. 166667 

5) 

.  041667 

6) 

. 008333 

7) 

.001389 

8) 

. 000198 

9) 

.  000025 

.  000003 

e      —       2.71828  (208) 

The  sixth  place  of  the  value  of  e  is  neglected  as  inaccurate. 

This  value  of  e  is  remarkable,  as  being  the  base  of  the  sys- 
tem of  logarithms  invented  by  Neper,  and  which  are  called  the 
Neperian  or  hyperbolic  logarithms. 

III.  The  values  of  R  (202)  and  e  (208)  give 

R  3437.7         _    3437.7    _ 

ToiTe  ~~  log.  (2.71828)  -  0.43429  ~  'y15'7'     (Z™> 

so  that  we  have  by  (199) 

Mer.  parts  of  L  =  7915.7  log.  co-tan.  (45°  —  |  L),     (210) 

which  agrees  with  the  explanation  of  Table  III.  given  in  the 
Preface  to  the  Navigator. 


<§>  46.]  mercator's  sailing.  97 

Correction  for  middle  latitude. 

45.    Examples. 

1.  Calculate  the  meridional  parts  for  latitude^45°  48'. 
Solution.  2)45°  48' 


45°  —  i  L  =  45°  --  22°  54'  =  22°  6' 
22°  6'     log.  cotan.         0.39141         log      9.59263 
7915.7  3.89849 


mer.  parts  of  45°  48' :=    3098  3.49112 

2.  Calculate  the  meridional  parts  of  latitude  28°  14'. 

Am.         1767. 

3.  Calculate  the  meridional  parts  of  latitude  83°  59/. 

Am.         10127. 

46.  Problem.  To  calculate  the  correction  for  middle 
latitude  sailing. 

Solution.  If  the  angle  DBC  (fig.  19.)  were  exactly  what 
it  should  be  in  order  that  the  hypothenuse  BD  should  be  the 
difference  of  longitude,  and  the  leg  JBCthe  departure,  it  would 
be  the  corrected  middle  latitude,  and  we  should  have 

diff.  long.  =sec.  cor.  mid.  lat.  X  departure 

z=  sec.  cor.  mid.  Iat.  X  diff.  lat.  X  tang,  course,  (211) 

which,  compared  with  (200)  gives,  by  dividing  by  tang,  course, 

mer.  diff.  lat.  =  sec.  cor.  mid.  lat.  X  diff.  lat.       (212) 

9 


98  NAVIGATION    AND    SURVEYING.  [CH.  V. 

Correction  for  middle  latitude. 

•  i   »  mer.  cliff,  lat.  «*.«* 

whence      sec.  cor.  mid.  lat.  =  — — — .  (213) 

diff.  lat.  ' 

If,  from  the  corrected  middle  latitude,  calculated  by 
this  formula,®  the  actual  middle  latitude  is  subtracted, 
the  correction  of  the  middle  latitude  is  obtained,  as  in 
the  table  on  p.  76  of  the  Navigator.  The  meridional 
difference  of  latitude  should  be  obtained  for  these  cal- 
culations, not  from  the  tables  of  meridional  parts,  but 
directly  from  the  tables  of  logarithmic  sines,  &c.  by 
means  of  (209)  ;  and  when  the  difference  of  latitude  is 
less  than  14°,  tables  should  be  used  in  which  the  loga- 
rithrris  are  given  to  seven  places  of  decimals. 

The  following  examples  are,  for  the  convenience  of  the 
learner,  limited  to  cases  for  which  the  common  tables  are 
sufficiently  accurate. 


47.     Examples. 

1.  Find  the  correction  for  middle  latitude  sailing,  when  the 
middle  latitude  is  35°,  and  the  difference  of  latitude  14°. 

Solution.  Greater  lat.  —  35°  +  7°  —  42° 

Less    lat.  —  35°  —  7°  =  28° 
45°  _  £  gr.  lat.  =  24°  cotan.  0.35142 
45°  —  J  less  lat.  =  31°  cotan.  0.22123 


0.13019 

log. 

9.11458 

7915.7 

3.89849 

diff.  lat.  =  840' 

ar.  co. 

7.07572 

corrected  mid.  lat.  =  35°  24'  sec.    10.08879 

correction         =  35°  24'  —  35°  =  24'. 


^  51.]  mercator's  sailing. 


99 


Cases  of  Mercator's  sailing. 


2.  Find  the  correction  for  middle  latitude  sailing,  when  the 
middle  latitude  is  60°,  and  the  difference  of  latitude  16°. 

Ans.     46'. 

3.  Find  the  correction  for  middle  latitude  sailing,  when  the 
middle  latitude  is  72°,  and  the  difference  of  latitude  20°. 

Ans.     124'. 

48.  Problem.  To  find  the  bearing  and  distance  of 
two  given  places.     [B.  p.  79.] 

Solution.    We  have  by  (fig.  20.)  for  the  bearing, 

diff.  long.  ..,.. 

tang,  bearing  =  ,._  ,  -   ,  (214) 

5  B         mer.  diff.  lat.  v        ' 

and  the  distance  is  found  by  (171). 

49.  Problem.  To  find  the  course,  distance,  and  dif- 
ference of  longitude,  when  both  latitudes  and  the  depar- 
ture are  given.     [B.  p.  80.] 

Solution.  The  course  is  found  by  (170),  the  difference  of 
longitude  by  (200),  and  the  distance  by  (171). 

50.  Problem.  To  find  the  distance  and  difference  of 
longitude,  when  both  latitudes  and  the  course  are  given. 
[B.  p.  82.] 

Solution.  The  distance  is  found  by  (171),  and  the  differ- 
ence of  longitude  by  (200). 

51.  Problem.  To  find  the  course  and  difference  of 
longitude,  when  both  latitudes  and  the  distance  are 
given.    [B.  p.  83.] 

Solution.  The  course  is  found  by  (174),  and  the  difference 
of  longitude  by  (200). 


100  NAVIGATION    AND    SURVEYING.  [cH.   V. 

Examples. 

52.  Problem.  To  find  the  distance,  the  difference  of 
latitude,  and  the  difference  of  longitude,  when  one  lati- 
tude, the  course,  and  departure  are  given,     [B.  p.  84.] 

Solution  The  distance  is  found  by  (177),  the  difference  of 
latitude  by  (176),  and  the  difference  of  longitude  by  (200). 

53.  Problem,  To  find  the  course,  the  difference  of  lati- 
tude, and  the  difference  of  longitude,  when  one  latitude, 
the  distance,  and  the  departure  are  given.     [B.  p.  85.] 

Solution,  The  course  is  found  by  (178),  the  difference  of 
latitude  by  (179),  and  the  difference  of  longitude  by  (200),  or 
by  the  following  proportion  deduced  from  the  similar  triangles 
of  (fig.  20.) 

Diff.  lat.  :  dep.  =z  mer.  diff.  lat.  :  diff.  long,  (215) 

54.    Examples. 

1.  A  ship  sails  from  Boston  a  distance  of  6747  miles,  upon  a 
course  S.  46°  59J'  E.  ;  to  find  the  place  at  which  it  arrives. 

Solution. 
Dist.  z=z  6747  3.82911 

Course  =  46°  59£'       cos.     9.83385  tang.  10.03022 


Diff.  lat.=76°  42' S.=4602/,  3.66296  mer.  d.  I.  =5007,  3.69958 
Lat.  Ieft  =  420-21'N.  mer.  p.  2810  


Lat  in  =  34°  21'  S.  2197  diff:  long.  =  5368'  3.72980 

=89°28/E. 

mer.  diff.  lat.  —    5007  long,  left  .=  71°   5'  W. 

long,  in         1829'  W. 


§  54.]  mercator's  sailing.  101 

Examples. 

Ans.     The  place  arrived  at  is  the  Cape  of  Good  Hope. 

2.    Find   the  bearing  and   distance   from   Moscow  to   St. 
Helena. 

Solution. 
Moscow,      lat.  55°  46'  N.     mer.  parts  4049       long.  37.°  33'  E. 
St.  Helena,  lat.  1 5°  55'  S.      mer.  parts     968       long.    5°  36' W. 


Diff.  lat.     —     71°  41/  mer.  diff.  lat.  5017  diff.l.  —  43°    9' 
af     4301'  =  2589' 

Mer. diff. lat.  =  5017  (ar.co.)    6.29956 
diff.  lon<r.—  2589  3.41313 


S.27°  18  W.  tang.  9.71269       sec.    10.05127 
diff.  lat.  ==  4301  3.63357 


dist.  z=  4840  miles        3.68484 

3.  A  ship  sails  from  a  position  200  miles  to  the  east  of  Cape 
Horn  a  distance  of  3635  miles,  upon  a  course  N.  N.E. ;  find 
the  position  at  which  it  has  arrived. 

Ans.  It  has  arrived  at  the  equator  in  the  longitude  of 
16°  6'  W. 

4.  Required  the  bearing  and  distance  of  Botany  Bay  from 
London. 

Ans.     The  Bearing     =     S.  57°  31'  E. 
Distance    z=       9551  miles. 

5.  A  ship  sails  northwesterly  from  Lima  until  it  arrives  in 
the  latitude  23°  7'  N.,  and  has  made  a  departure  of  9983 
miles ;  find  the  place  at  which  it  has  arrived. 

Ans.     Canton. 
9* 


102  NAVIGATION    AND    SURVEYING.  [CH.   V. 

Examples. 

6.  A  ship  sails  from  Disappointment  Island  in  the  North 
Pacific  Ocean,  upon  a  course  S.  61°  41'  E.,  until  it  has  arrived 
in  latitude  14°  7'S. ;  find  the  place  at  which  it  has  arrived. 

Ans.     The  Disappointment  Islands  in  the  South  Pacific 
Ocean. 

7.  A  ship  sails  from  Icy  Cape  (North  West  Coast  of  Amer- 
ica) a  distance  of  9138  miles  southeasterly,  when  it  has  arrived 
in  latitude  62°  30'  S. ;  find  the  place  at  which  it  has  arrived. 

Ans.     Yankee  Straits  in  New  South  Shetland. 

8.  A  ship  sails  from  Java  Head,  upon  a  course  S.  68°  53' W  , 
until  it  has  made  a  departure  of  4749  miles  ;  find  the  position 
at  which  it  has  arrived. 

Ans.     It  has  arrived  at  a  position  180  miles  south  of  the 
Cape  of  Good  Hope. 

9.  A  ship  sails  southeasterly  from  the  South  Point  of  the 
Great  Bank  of  Newfoundland  a  distance  of  2821  miles,  when 
it  has  made  a  departure  of  910  miles  ;  find  the  position  at 
which  it  has  arrived. 

Ans.     Its  position  is  208  miles  north  of  Cape  St.  Roque. 


§  56.]  SURVEYING. 


Area  of  triangle. 


CHAPTER    VI. 


SURVEYING. 


55.  The  object  of  Surveying  is  to  determine  the 
dimensions  and  areas  of  portions  of  the  earth's  surface. 
In  the  application  of  Plane  Trigonometry,  the  portions 
of  the  earth  are  supposed  to  be  so  small  that  the  curva- 
ture of  the  earth  is  neglected.  They  are,  in  this  case, 
nothing  more  than  common,  fields  bounded  by  lines 
either  straight  or  curved. 

56.  Problem.  To  find  the  area  of  a  triangular  fields 
when  its  angles  and  one  of  its  sides  are  known. 

Solution.  Let  ABC  (fig.  2.)  be  the  triangle  to  be  measured, 
and  c  the  given  side.  The  area  of  the  triangle  is  equal  to 
half  the  product  of  its  base  by  its  altitude,  or 

(216) 


area 

of     ABC=$bp. 

But,  by  (123), 

sin 

.  b  :  sin.  B  :  :  c  :  6, 

whence 

c  sin. B 

~    sin.  C  ' 

and,  by  (124), 

p  =z  c  sin.  A. 

104  NAVIGATION    AND    SURVEYING.  [CH.  VI. 

Area  of  triangle. 


Substituting  (216),  we  have 

~  2  sin.  C 


c   AT>n       c2  sin.  ^4  sin.  JB  #««*% 

area  of  ABC  =  -___-.  (217) 


57.  Problem.  To  find  the  area  of  a  triangular  field, 
when  two  of  its  sides  and  the  included  angle  are 
known. 

Solution.  Let  ABC  (fig.  2.)  be  the  triangle  to  be  measured, 
6  and  c  the  given  sides,  and  A  the  given  angle.  Then,  by 
(216), 

area  of  ABC  —  £  b  p, 

and,  by  (124), 

p  zz:  c  sin.  A. 
Hence 

area  of  ABC  =  }  b  c  sin.  A.  (218) 

or,  the  area  of  a  triangle  is  equal  to  half  the  continued 
product  of  two  of  its  sides  and  the  sine  of  the  included 
angle. 

58.  Problem.  To  find  the  area  of  a  triangular  field, 
when  its  three  sides  are  known. 

Solution.  Let  ABC  (fig.  1.)  be  the  given  triangle.  Then, 
by  (218), 

area  of  ABC  z=z  £  b  c  sin.  A  ; 
but,  by  (151), 

sin.  A  =  iW['(«-*)('-*)  ('-«)] 
b  c 

in  which  s  denotes  the  half  sum  of  the  three  sides  of  the  tri- 
angle. 


§  59.]  SURVEYING.  105 

Area  of  triangle. 

Hence 

b  c  sin.  A  z=  2  a/[s  (s  —  a)  (s  —  b)  (s  —  c)]  ; 
and 

area  of  ABC  =  s/[s  (s  —  a)  (s  —  b)  (s  —  c)]  ;    (219) 

or,  to  find  the  area  of  a  triangular  field,  subtract  each 
side  separately  from  the  half  stun  of  the  sides  ;  and  the 
square  7%oot  of  the  continued  'product  of  the  half  sum 
and  the  three  remainders  is  the  required  area. 

59.    Examples. 

1.  Given  the  three  sides  of  a  triangular  field,  equal  to  45.56 
ch.,  52.98  ch.,  and  61.22  ch.  ;  to  find  its  area. 

Solution.    In  (fig.  1.)  let  a  =  45.56  ch.,   b  =1  52.98  ch., 
c  =  61.22  ch. 

2  5=  159.76  ch. 

s  z=    79.88  ch.  1.90244 

s  —  a  =    34.32  ch.  1.53555 

s  —  b=    26.90  ch.  1.42975 

5  —  c=z    18.66  ch.  1.27091 


6.13865 


Area  of  ABC—  1 173.  1  sq.  ch.     3.06932 

Ans.     The  area  =  117  A.  1  R.  9  r. 

2.    Given  the  three  sides  of  a  triangular  field  equal  to  32.56 
ch.,  57.84  ch.,  and  44.44  ch.  ;  to  find  its  area. 

Ans.     The  area  =  71  A.  3  R.  29  r. 


106  NAVIGATION    AND    SURVEYING.  [CH.  VI. 

Area  of  rectilinear  field. 

3.  Given  one  side  of  a  triangular  field  equal  to  17.35  ch., 
and  the  adjacent  angles  equal  to  100°  and  70°  ;  to  find  its 
area. 

Ans.     The  area  =  85  A.  3  R.  16  r. 

4.  Given  two  sides  of  a  triangular  field  equal  to  12.34  ch. 
and  17.97  ch.,  and  the  included  angle  equal  to  44°  56' ;  to 
find  its  area. 

Ans.     The  area  =  7  A.  3  R.  13  r. 

60.  Problem.  To  find  the  area  of  an  irregular  field 
bounded  by  straight  lines. 

First  Method  of  Solution.  Divide  the  field  into 
triangles  in  any  manner  best  suited  to  the  nature  of  the 
ground.  Measure  all  those  sides  and  angles  which  can 
be  measured  conveniently,  remembering  that  three 
parts  of  each  triangle,  one  of  which  is  a  side,  must  be 
known  to  determine  it. 

But  it  is  desirable  to  measure  more  than  three  parts  of  each 
triangle,  when  it  can  be  done ;  because  the  comparison  of 
them  with  each  other  will  often  serve  to  correct  the  errors  of 
observation.  Thus,  if  the  three  angles  were  measured,  and 
their  sum  found  to  differ  from  180°,  it  would  show  there  was 
an  error ;  and  the  error,  if  small,  might  be  divided  between 
the  angles;  but  if  the  error  was  large,  it  would  show  the  ob- 
servations were  inaccurate,  and  must  be  taken  again. 

The  area  of  each  triangle  is  to  be  calculated  by  one 
of  the  preceding  formulas,  and  the  sum  of  the  areas  of 
the  triangles  is  the  area  of  the  whole  field. 

This  method  of  solution  is  general,  and  may  be  ap- 
plied to  surfaces  of  any  extent,  provided  each  triangle 


<§>  60.]  SURVEYING.  107 

Rectangular  surveying. 

is  so  small  as  not  to  be  affected   by  the  earth's  curva- 
ture. 

Second  Method  of  Solution.  Let  ABCEFH  (fig.  21.)  be 
the  field  to  be  measured.  Starting  from  its  most  easterly  or  its 
most  westerly  point,  the  point  A  for  instance,  measure  succes- 
sively round  the  field  the  bearings  and  lengths  of  all  its  sides. 

Through  A  draw  the  meridian  NS,  on  which  let  fall  the 
perpendiculars  BB,  CO,  EE ",  FF',  and  HH'.  Also  draw 
CB'E',  EF",  and  HF"  parallel  to  NS. 

Then  the  area  of  the  required  field  is 

ABCEFH  =z  AC'CEFF'  —  [AC'CB  +  AHFF1]. 

But 

AC'CEFF'  =  C'CEE'  +  E'EFF' ; 

and 

AC'CB  +  AHFF'  =  C'CBB'  +  BBA  +  AHH' 
+  HHFF1. 
Hence 

ABCEFH  =  [C'CEE'  +  EEFF]  —  [C'CBB' 

+  BBA  +  AHH'  +  H'HFF]  ; 

or  doubling  and  changing  a  very  little  the  order  of  the  terms, 

2  ABCEFH  —  [2  C'CEE  +  2  E'EFF']  —       £ 
[2  BBA  +  2  CC'BB'  +  2  HHFF'  +  2  AHH'].  >  (22°} 

Again, 

2  5,jB4       =    BB'  X  -4B' 

2  CC'BB'   =z  {BB'  +  CC)  X  B'C 

2  C'CjEjET   =  (EE'  +  CC)  X  EC"  . 

2  EEFF'  =  (EE'  +  JLF'J  X  JB'*"  ^ 

2  HHFF'  =  (Hi/'  +  JFF")  X  i?7^' 

2  jUTtf'      =  i?i/'  X  -4H7. 


108 


NAVIGATION    AND    SURVEYING. 


[CH.  VI. 


Rectangular  surveying. 


So  that  the  determination  of  the  required  area  is  now  reduced 
to  the  calculation  of  the  several  lines  in  the  second  members 
of  (221.)  But  the  rest  of  the  solution  may  be  more  easily 
comprehended  by  means  of  the  following  table,  which*is  pre- 
cisely similar  in  its  arrangement  to  the  table  actually  used  by 
surveyors,  when  calculating  areas  by  this  process. 


Sides 
AB 

N. 

s. 

E. 

W. 

Dep. 

Sum. 

N. Areas. 

S.  Areas. 

AB 

BB' 

BB' 

BB' 

BB'A 

BC 

BC 

BB' 

CC 

BB'+CC 

CC'BB' 

CE 

CE' 

EE' 

EE' 

CC'  +  EE' 

CCEE' 

EF 

EF' 

pp// 

FF' 

EE'+FF' 

EEFF' 

FH 

FH' 

FF7" 

HH' 

FF'-f-  HH' 

H'HFF' 

HA 

HA 

Hir 

O 

HH' 

AHH' 

In  the  first  column  of  the  table  are  the  successive 
sides  of  the  field. 

In  the  second  and  third  columns  are  the  differences 
of  latitude  of  the  several  sides,  the  column  headed  N, 
corresponding  to  the  sides  running  in  a  northerly  direc- 
tion, and  that  headed  S,  corresponding  to  those  running 
in  a  southerly  direction. 

These  two  columns  are  calculated  by  the  formula 
Diff.  lat.  —  dist.  X  cos.  bearing. 

In  the  fourth  and  fifth  columns  are  the  departures  of 
the  several  sides ;  the  column  headed  E,  corresponding 
to  the  sides  running  in  an  easterly  direction,  and  that 
headed  W,  to  those  running  in  a  westerly  direction. 


§  60.]  SURVEYING.  109 

Rectangular  survey. 

These  two  columns  are  calculated  by  the  formula 
Departure  s=  dist.  X  sin.  bearing. 

In  the  sixth  column,  headed  Departure,  are  the  depar- 
tures of  the  several  vertices,  which  end  each  side  of 
the  field  from  the  vertex  A.  This  column  is  calculated 
from  the  two  columns  E  and  W,  in  the  following  man- 
ner. The  first  number  in  column  Departure  is  the 
same  as  the  first  in  the  two  columns  E  and  W;  and 
every  other  number  in  column  Departure  is  obtained  by 
adding  the  corresponding  number  in  columns  E  and 
W,  if  it  is  of  the  same  column  with  the  first  number 
in  those  two  columns,  to  the  previous  number  in  column 
Departure  ;  and  by  subtracting  it,  if  it  is  of  a  different 
column. 

Thus, 

BB'  ==  BB> 

CC  ==  BB"  =2  BB'  —  BB" 

EE1  3=  E'E"  +  EE"  =.  CC  +  EE" 
FF'  =  F'F"  +  FF" '  =  EE  +  FF" 

HH  =:  F'F'"  =  FF' FF'" 

O    =:  HH'  —  HH'. 

In  the  seventh  column,  headed  Sum,  are  the  first 
factors  of  the  second  members  of  (221).  This  column 
is  calculated  from  column  Departure  in  the  following 
manner.  The  first  number  in  column  Sum  is  the  same 
as  the  first  in  column  Departure ;  every  other  number 
in  column  Sum  is  the  sum  of  the  corresponding  num- 
10 


110  NAVIGATION    AND    SURVEYING.  [CH.  VI. 

Rectangular  survey. 

ber  in  column  Departure  added  to  the  previous  number 
in  column  Departure,  as  is  evident  from  simple  inspec- 
tion. 

In  the  eighth  and  ninth  columns  are  the  values  of 
the  areas,  which  compose  the  first  members  of  (221). 
These  columns  are  calculated  by  multiplying  the  mem- 
bers in  column  Sum  by  the  corresponding  numbers  in 
columns  N  and  S,  which  contain  the  second  factors  of 
the  second  members  of  (221).  The  products  are  writ- 
ten in  the  column  of  North  Areas,  when  the  second  fac- 
tors are  taken  from  column  N,  and  in  that  of  South 
Areas,  when  the  second  factors  are  taken  from  col- 
umn S. 

If  we  compare  the  columns  of  North  and  South 
Areas  with  (220),  we  find  that  all  those  areas,  which 
are  preceded  by  the  negative  sign,  are  the  same  with 
those  in  the  column  of  North  Areas  ;  while  all  those, 
which  are  connected  with  the  positive  sign,  belong  to 
the  column  of  South  Areas.  To  obtain,  therefore,  the 
value  of  the  second  member  of  (220),  that  is,  of  double 
the  required  area,  we  have  only  to  find  the  difference 
between  the  sums  of  the  columns  of  North  and  South 
Areas.     [B.  p.  107.] 

61.  Corollary.  The  columns  N,  S,  E,  and  W,  are  those 
which  would  be  calculated  in  Traverse  Sailing,  if  a  ship  was 
supposed  to  start  from  the  point  A,  and  proceed  round  the 
sides  of  the  field  till  it  returned  to  the  point  A.  The  differ- 
ence of  the  sums  of  columns  N  and  S  is,  then,   by  traverse 


§  62.]  SURVEYING.  ill. 


Correction  of  errors. 


sailing,  the  difference  of  latitude  between  the  point  from  which 
the  ship  starts,  and  the  point  at  which  it  arrives;  and  the 
difference  of  columns  E  and  W  is  the  departure  of  the  same 
two  points.  But  as  both  the  points  are  here  the  same,  their 
difference  of  latitude  and  their  departure  must  be  nothing,  or 

Sum  of  column  N  =  sum  of  column  S  ; 

Sum  of  column  E  =  sum  of  column  W. 

But  when,  as  is  almost  always  the  case,  the  sums  of  these 
columns  differ  from  each  other,  the  difference  must  arise  from 
errors  of  observation.  If  the  error  is  great,  new  observations 
must  be  taken  ;  but  if  it  is  small,  it  may  be  divided  among 
the  sides  by  the  following  proportion. 

The  sum  of  the  sides  :  each  side  =  whole  error  : 
error  corresponding  to  each  side. 

The  errors  corresponding  to  the  sides  are  then  to  be 
subtracted  from  the  differences  of  latitude,  or  the  de- 
partures which  are  in  the  larger  column,  and  added  to 
to  those  which  are  in  the  smaller  column. 


62.    Examples. 

1.  Given  the  bearings  and  lengths  of  the  sides  of  a  field,  as 
in  the  three  first  columns  of  the  following  table ;  to  find  its 
area. 


Solution.    The  table  is  computed  by  §  60. 


112 


NAVIGATION    AND    SURVEYING. 


[CH.  VI. 


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*63.] 


SURVEYING. 


113 


Area  of  irregular  field. 


2.  Given  the  lengths  and  bearings  of  the  sides  of  a  field,  as 
in  the  following  table  ;  to  find  its  area. 


No. 

Bearings. 

Dist. 

1 

N.  17°  E. 

25  ch. 

2 

East. 

28  ch. 

3 

South. 

54  ch. 

4 

S.  4°  W. 

22  ch. 

5 

N.  33°  W. 

62  ch. 

Ans.     The  area  =  173  A.  0  R.  36  r. 


63.  Problem.  To  find  the  area  of  a  field  bounded  by 
sides,  irregularly  curved. 

Solution.  Let  ABCEFHIKL  (fig.  22.)  be  the  field  to  be 
measured,  the  boundary  ABCEFHIKL  being  irregularly 
curved.  Take  any  points  C  and  F,  so  that  by  joining  AC, 
CF,  and  FL,  the  field  ACFLy  bounded  by  straight  lines,  may 
not  differ  much  from  the  given  field. 

Find  the  area  of  ACFL,  by  either  of  the  preceding  meth- 
ods, and  then  measure  the  parts  included  between  the  curved 
and  the  straight  sides  by  the  following  method  of  offsetts. 

Take  the  points  a,  6,  c,  d,  so  that  the  lines  A  a,  a  b,  b  c, 
cdy  dC  may  be  sensibly  straight.  Let  fall  on  AC  the  per- 
pendiculars a  a1,  bb1,  cc't  dd'.  Measure  these  perpendiculars, 
and  also  the  distances  Aa'9  b'c',  b'c'9  c'd',  d'C. 

10* 


114  NAVIGATION    AND    SURVEYING.  [cH.  VI. 

Area  of  irregular  field. 

The  triangles  Acta',  Cdd1,  and  the  trapeziums  aba'b', 
bcb'c',  cdc'd  are  then  easily  calculated,  and  their  sum  is 
the  area  of  ABC. 

In  the  same  way  may  the  areas  of  CEF,  FHI,  and  IKL 

be  calculated ;  and  then  the  required  area  is  found  by  the 
equation 

ABCEFH1KL  =s  ACFL  —  ABC  +  CEF  + 
FHI  —  IKL. 

Example. 

Given  (fig.  22.)  A  a'  =  5  ch.,  a'b'  =  2  ch.,  b1  c'  =  6  ch., 
e'tf  =  1  ch.,  d'C  =  4  ch.  ;  also  a  a'  £s  3  ch.,  bb'  =  2  ch.} 
cc'  =  25  ch.,  rfrf'  z=  1  ch. ;  to  find  the  area  of  ABC. 

Arts.     Required  area  z:2A.  3  R.  36  r. 


§  67.]  HEIGHTS    AND    DISTANCES.  115 

Horizon.  Bearing. 


CHAPTER   VII. 


HEIGHTS    AND    DISTANCES. 


64.  The  plane  of  the  sensible  horizon  at  any  place, 
is  the  tangent  plane  to  the  earth's  surface  at  that  place. 
[B.  p.  48.] 

The  horizontal  plane  coincides  with  that  of  the  surface  of 
tranquil  waters,  when  this  surface  is  so  small  that  its  curvature 
may  be  neglected  ;  and  it  is  perpendicular  to  the  plumb  line. 

65.  The  angle  of  elevation  of  an  object  is  the  angle 
which  the  line  drawn  to  it  makes  with  the  horizontal 
plane,  when  the  object  is  above  the  horizon ;  the  angle 
of  depression  is  the  same  angle  when  the  object  is  be- 
low the  horizon. 

66.  The  bearing  of  one  object  from  another  is  the 
angle  included  by  the  two  lines  which  are  drawn  from 
the  observer  to  these  two  objects. 

67.  Problem.  To  determine  the  height  of  a  vertical 
tower,  situated  on  a  horizontal  plane.     [B.  p.  94.] 

Solution.  Observation.  Let  AB  (fig.  23.)  be  the  tower, 
whose  height  is  to  be  determined.  Measure  off  the  distance 
BC  on  the  horizontal  plane  of  any  convenient  length.  At 
the  point  C  observe  the  angle  of  elevation  ACB. 


116  NAVIGATION    AND    SURVEYING.  [CH.   VII. 

Height  of  vertical  tower. 

Calculation.  We  have,  then,  given  in  the  right  triangle 
ACB  the  angle  C  and  the  base  BC,  as  in  problem,  §  33  of  PL 
Trig.,  and  the  leg  AB  is  found  by  (22). 

Example. 

At  the  distance  of  95  feet  from  a  tower,  the  angle  of  eleva- 
tion of  the  tower  is  found  to  be  48°  19'.  Required  the  height 
of  the  tower. 

Arts.     106.69  feet. 

68.  Problem.  To  find  the  height  of  a  vertical  tower 
situated  on  an  inclined  plane. 

Solution.  Observation.  Let  AB  (fig.  24.)  be  the  tower 
situated  on  the  inclined  plane  BC.  Observe  the  angle  B, 
which  the  tower  makes  with  the  plane.  Measure  off  the  dis- 
tance BC  of  any  convenient  length.  Observe  the  angle  C, 
made  by  a  line  drawn  to  the  top  of  the  tower  with  BC. 

Calcidation.  In  the  oblique  triangle  ABC,  there  are  given 
the  side  BC  and  the  two  adjacent  angles  B  and  C,  as  in 
§  72  of  PL  Trig. 

Example. 

Given  (fig.  24.)  BC '=  89  feet,  B  =  113°  12',  C "=  23° 
27' ;  to  find  AB. 

Ans.     AB  =  51.595  feet. 

69.  Problem.  To  find  the  distance  of  an  inaccessible 
object.    [B.  p.  89  and  95.] 

Solution,  Observation.  Let  B  (fig.  2.)  be  the  point,  the  dis- 
tance of  which  is  to  be  determined,  and  A  the  place  of  the 


<§>  72.]  HEIGHTS    AND    DISTANCES.  117 

Distance  of  inaccessible  object. 

observer.     Measure  off  the  distance  AC  of  any  convenient 
length,  and  observe  the  angles  A  and  C. 

Calculation.    AB  and  BC  are  found  by  §  72  of  PL  Trig. 

70.  Corollary.  The  perpendicular  distance  BP  of  the 
point  B  from  the  line  AC,  and  the  distances  AP  and  PC 
are  found  in  the  triangle  ABP  and  BPC,  by  §  31  of  PI.  Trig. 

71.  Corollary.  Instead  of  directly  observing  the  angles  A 
and  C,  the  bearings  of  the  lines  ABy  AC,  and  BC  may  be 
observed,  when  the  plane  ABC  is  horizontal,  and  the  angles 
A  and  C  are  easily  determined. 


72.     Examples. 

1.  An  observer  sees  a  cape,  which  bears  N.  by  E. ;  after 
sailing  30  miles  N.  W.  he  sees  the  same  cape  bearing  east ; 
find  the  distance  of  the  cape  from  the  the  two  points  of  obser- 
vation. 

Arts.     The  first  distance  ~  21.63  miles. 
The  second  dist.    —  25.43  miles. 

2.  Two  observers  stationed  on  opposite  sides  of  a  cloud  ob- 
serve the  angles  of  elevation  to  be  44°  56',  and  36°  4',  their 
distance  apart  being  700  feet ;  find  the  distance  of  the  cloud 
from  each  observer  and  its  perpendicular  altitude. 

Ans.  Distances  from  observers  =:  417.2  feet,  and  =  500.6  ft. 

Height  =z  294.7  feet. 

3.  The  angle  of  elevation  of  the  top  of  a  tower  at  one  sta- 
tion is  observed  to  be  68°  19',  and  at  another  station  546  feet 


118  NAVIGATION    AND    SURVEYING.  [CH.  VII. 

Height  of  inaccessible  object. 

farther  from  the  tower,  the  angle  of  elevation  is  32°  34';  find 
the  height  and  distance  of  the  tower,  the  two  points  of  obser- 
vation being  supposed  to  be  in  the  same  horizontal  plane  with 
the  foot  of  the  tower. 

Ans.     The  height  ;         .         .         —  234.28  ft. 

The  dist.  from  the  nearest  point  of  observ.  =  135.86  ft. 

73.  Problem*  To  find  the  distance  of  an  object  from 
the  foot  of  a  tower  of  known  height,  the  observer  being 
at  the  top  of  the  tower. 

Solution.  Observation.  Let  the  tower  be  AB  (fig.  23.),  and 
the  object  C.     Measure  the  angle  of  depression  HAC. 

Calculation.     Since 

ACB  —  HAC, 

we  know  in  the  triangle  ACB  the  leg  AB  and  the  opposite 
angle  C,  as  in  §  32  of  PL  Trig. 

Example. 

Given  the  height  of  the  tower  —  150  feet,  and  the  angle  of 

depression  ==17°  25' ;  to  find  the  distance  from  the  foot  of 
the  tower. 

Ans.  478=16  feet. 

74.  Problem.  To  find  the  height  of  an  inaccessible 
object  above  a  horizontal  plane,  and  its  distance  from 
the  observer.    [B.  p.  96.] 

Solution.  Observation.  Let  A  (fig.  25.)  be  the  object.  At 
two  different  stations,  B  and  C,  whose   distance  apart  and 


§  75.]  HEIGHTS    AND    DISTANCES.  119 


Distance  of  two  objects. 


bearing  from  each  other  are  known,  observe  the  bearings  of 
the  object,  and  also  the  angle  of  elevation  at  one  of  the  stations, 
as  B. 

Calculation.  In  the  triangle  BCD,  the  side  BC  and  its 
adjacent  angles  are  known,  so  that  BD  is  found  by  §  72  of  PI. 
Trig.  In  the  right  triangle  ABD,  the  height  AD  is,  then, 
computed  by  §  33  of  PI.  Trig. 

Example. 

At  one  station  the  bearing  of  a  cloud  is  N.N.W.,  and  its 
angle  of  elevation  50°  35'.  At  a  second  station,  whose  bearing 
from  the  first  station  is  N.  by  E.,  and  distance  5000  feet,  the 
bearing  of  the  cloud  is  W.  by  N.  ;  find  the  height  of  the 
cloud. 

Ans.     7316.5  feet. 

75.  Problem.  To  find  the  distance  of  two  objects, 
ivhose  relative  position  is  known.     [B.  p.  90.] 

Solution.  Observation.  Let  B  and  C  (fig.  1.)  be  the  two 
known  objects,  and  A  the  position  of  the  observer.  Observe 
the  bearings  of  B  and  C  from  A, 

Calculation.  In  the  triangle  ABC,  the  side  BC  and  the 
two  angles  are  known.  The  sides  of  AB  and  AC  are  found 
by  §  72  of  PI.  Trig. 

Example. 

The  bearings  of  the  two  objects  are,  of  the  first  N.  E.  by 
E.,  and  of  the  second  E.  by^S. ;  the  known  distance  of  the  first 
object  from  the  second  is  23.25  miles,  and  the  bearing  N.  W.  ; 
find  their  distance  from  the  observer. 


120  NAVIGATION    AND    SURVEYING.  [CH.  VII. 

Distance  apart  of  two  objects. 

Ans.     The  distance  of  the  first  object  is  zn  18.27  miles. 
That  of  the  second  object  =  32.25  miles. 

76.  Problem.  To  find  the  distance  apart  of  two  ob- 
jects separated  by  an  impassable  barrier.    [B.  p.  91.] 

Solution,  Observation.  Let  A  and  B  (fig.  1.)  be  the  ob- 
jects ;  the  distance  of  which  from  each  other  is  sought. 
Measure  the  distances  and  bearings  from  any  point  C  to  both 
A  and  B. 

Calculation.  In  the  triangle  ABC  the  two  sides  A  C  and 
BC  and  the  included  angle  C  are  known.  The  sides  AB 
and  BC  may  be  found  by  §  81  of  PI.  Trig. 

Example. 

Two  ships  sail  from  the  same  port,  the  one  N.  10°  E.  a  dis- 
tance of  200  miles,  the  second  N.  70°  E.  a  distance  of  150 
miles;  find  their  bearing  and  distance. 

Ans.     The  distance  =  180.3  miles. 

The  bearing  of  the  first  ship  from  the  second  =  N.  36°  6'  W. 

77.  Problem.  To  find  the  distance  apart  of  two  in- 
accessible objects  situated  in  the  same  plane  with  the 
observer,  and  their  bearing  from  each  other.  [B.  p.  92.] 

Solution.  Observation.  Let  A  and  B  (fig.  26.)  be  the  two 
inaccessible  objects.  At  two  stations,  C  and  E,  observe  the 
bearings  of  A  and  B  and  the  bearing  and  distance  of  C  from  E. 

Calculation.  In  the  triangle  AEC  we  have  the  side  CE, 
and  the  angles  ACE  and  AEC9  so  that  AC  is  found  by  §  72 
of  PL  Trig. 


§  78.]  HEIGHTS   AND    DISTANCES.  121 

Distance  apart  of  two  objects. 


In  the  same  way  B  C  is  calculated  fronvthe  triangle  BCE. 

Lastly,  in  triangle  ABC,  we  know  the  two  sides  AC  and  . 
BC,  and  the  included  angle  for 

ACB  =  ACE  —  BCE. 

Hence  AB  and  the  angles  BAC  and  CBA  are  found  by 
§81. 

Example. 

An  observer  from  a  ship  saw  two  headlands ;  the  first  bore 
E.  N.  E.,  and  the  second  N.  W.  by  N.  After  he  had  sailed 
N.  by  W.  16.25  miles,  the  first  headland  bore  E.  and  the 
second  N.  W.  by  W. ;  find  the  bearing  and  distance  of  the 
first  headland  from  the  second. 

Ans.     Distance  =  55.9  miles. 

Bearing    —  N.  65°  33'  W. 

78.  Problem.  To  find  the  distance  of  an  object  of 
known  height,  which  is  just  seen  in  the  horizon. 

Solution.  I.  If  light  moved  in  a  straight  line,  and  if  A 
(fig.  27.)  were  the  eye  of  the  observer,  and  B  the  object,  the 
the  straight  line  APB  would  be  that  of  the  visual  ray.  The 
point  P,  at  which  the  ray  touches  the  curved  surface  CPD  of 
the  earth,  is  the  point  of  the  visible  horizon  at  which  the  ob- 
ject is  seen.  The  distances  PA  and  PB  may  be  calculated 
separately,  when  the  heights  AC  and  BD  are  known.  For 
this  purpose,  let  O  be  the  earth's  centre,  let  BD  be  produced 
to  E,  and  let 

h  =i  AC,     H=z  BD, 

I   =  PA,     L  =  PB, 

R  z=z  the  earth's  radius 
11 


122  NAVIGATION    AND    SURVEYING.  [CH*  VII. 

Distance  of  an  object  seen  in  the  horizon. 

Since  BP  is  a  tangent,  and  BOE  a  secant  to  the  earth, 
we  have 

BE:  BP  =  BP  :  BD; 

and  BD  is  so  small  in  comparison  with  the  radius,  that  we 
may  take 

BE=  DE  =  2R, 


and  the  above  proportion  becomes 

2  R  :  L  =  L  :  H ; 

whence 

2,2  —  2  RH,     L  &  V(2  RH)> 

(222) 

H  -    L2 
H  -21V 

(223) 

and  in  the  same  way 

72  =  2  R  h,       I—  \/(2  R  h), 

(224) 

7            l2 

(225) 

II.  Light  does  not,  however,  move  in  a  straight  line  near 
the  earth's  surface,  but  in  a  line  curved  towards  the  other 
centre ,  which  is  nearly  an  arc  of  a  circle,  whose  radius  is 
seven  times  the  earth's  radius ;  so  that  for  the  point  of  con- 
tact P  and  the  distances  I  and  L,  the  positions  of  the  eye  and 
of  the  object  are  A'  and  B'.     Now  if  we  put 

BB'  =  H>,  BD  =  Hx  —  H—  H 

A!C—hx, 

we   can   find   the   value  of  H1  with   sufficient   accuracy  by 
changing  in  (223)  R  into  7  R,  which  gives 


§  79.]  HEIGHTS    AND    DISTANCES.  123 

Distance  of  an  object  seen  in  the  horizon. 


L2 

H'  —            —  1  H 

Hx  —H  —  H=^H= 

SL2 
"  7R' 

(226) 

L^^dRH,). 

(227) 

whence 


III.  In  calculating  the  value  of  L  by  (227),  it  is  usually 
desired  in  statute  miles,  while  the  height  H1  is  given  in  feet. 
Now  we  have  in  the  Preface  to  the  Navigator,  page  v, 

R  z=  20911790  feet,  (228) 

whence  £  R  =  48794177  feet, 

log.  */(i  R)  =  i  log.  i  R  =  3.84418, 

and 

log.  (L  in  feet)  =  3.84418  +  %  log.  (H1  in  feet). 

„  -  .        .,  L  in  feet 

But  L  m  miles  =    ^       , 

so  that 

log.  L  in  miles  =  log.  L  in  feet  —  3.72263 

=  0.12155  +  £  log.  Hx  in  feet,     (229) 

which  agrees  with  the  formula  given  in  the  preface  to  the 
Navigator  for  calculating  table  X. 

IV.  The  Table  may  be  used  for  finding  L  or  7,  when  H1 
and  h  1  are  given,  and  then  the  required  distance  is  the  sum 
of  L  and  I. 

79.  Corollary.  Table  X  gives  the  correction  for  the  error 
which  is  committed  in  §  67/-  by  neglecting  the  earth's  curva- 


124  NAVIGATION    AND    SURVEYING.  [CH.  VII. 

Table  X  of  the  Navigator. 


ture,  for  it  is  evident  that  to  the  height  BP  (fig.  28.)  of  the 
object  above  the  visible  level  must  be  added  the  height  PC  of 
the  level  above  the  curved  surface  of  the  earth,  as  in  B.  p.  95. 


80.    Examples. 

1.  Calculate  the  distance  in  table  X  at  which  an  object  can 
be  seen  from  the  surface  of  the  earth,  when  its  height  is  5000 
feet. 

Solution. 

J  log.  5000  ~  £  (3.69897)  =  1.84948 

constant  log.  =:  0.12155 


dist.  =  93.5  (as  in  table  X)  1.97103 

2.  Being  on  a  hill  200  feet  above  the  sea,  I  see  just  appear- 
ing in  the  horizon  the  top  of  a  mast,  which  I  know  to  be  150 
feet  above  the  water  ;  how  far  distant  is  it  ? 

Solution.    By  table  X, 

200  feet  corresponds  to  18.71  miles. 

150  feet  corresponds  to  16.20 


The  distance  is  34.91  miles. 

3.  At  the  distance  of  7J  statute  miles  from  a  hill  the  angle 
of  elevation  of  its  top  is  2°  13' ;  find  its  height  in  feet,  the 
observer  being  20  feet  above  the  sea. 


§  80.]  HEIGHTS    AND    DISTANCES.  125 

Distance  of  an  object  seen  in  the  horizon. 

Solution. 

2°  13  tang.  8.58779 

7£  miles  =  39600  4.59770  By  table  X. 


1533  feet  3.18549  7.50 

1  foot  correction,       height  20  5.12 

height  =  1534  feet,  height    1  1.58 

4.  Calculate  the  distance  in  table  X,  when  the  height  is 
450  feet. 

Ans.     28.06  miles. 

5.  Upon  a  height  of  5000  feet,  the  top  of  a  hill,  one  mile 
high,Jris  just  visible  in  the  horizon ;  how  far  distant  is  the 
hill? 

Ans.     189.6  miles. 

6.  At  the  distance  of  25  miles  from  a  mountain  the  angle 
of  elevation  of  its  top  is  3° ;  find  its  height,  the  observer  being 
60  feet  above  the  intervening  sea. 

Ans.     7033  feet. 


11* 


SPHERICAL    TRIGONOMETRY. 


SPHERICAL    TRIGONOMETRY. 


CHAPTER  I. 


DEFINITIONS. 


1.  Spherical  Trigonometry  treats  of  the  solution  of 
spherical  triangles. 

A  Spherical  Triangle  is  a  portion  of  the  surface  of  a 
sphere  included  between  three  arcs  of  great  circles. 

In  the  present  treatise  those  spherical  triangles  only  are 
treated  of,  in  which  the  sides  and  angles  are  less  than  180°. 

2.  The  angle,  formed  by  two  sides  of  a  spherical  tri- 
angle, is  the  same  as  the  angle  formed  by  their  planes. 

3.  Besides  the  usual  method  of  denoting  sides  and 
angles  by  degrees,  minutes,  &c. ;  another  method  of 
denoting  them  is  so  often  used  in  Spherical  Astronomy, 
that  it  will  be  found  convenient  to  explain  it  here. 

The  circumference  is  supposed  to  be  divided  into  24 
equal  arcs,  called  hours ;  each  hour  is  divided  into  60 
minutes  of  time,  each  minute  into  60  seconds  of  time, 
and  so  on. 

Hours,  minutes,  seconds,  &c.  of  time  are  denoted  by  h,  ??i, 
s,  &c. 


130  SPHERICAL  TRIGONOMETRY.         [CH.  I. 

Arcs  expressed  in  time. 

4.  Problem.    To  convert  degrees,  minutes,  fyc.  into 
hours,  minutes,  fyc.  of  time. 

Solution.    Since 

360°  =  24 h 

we  have  15°  =  l\  and  1°  =  T^h  =  4W, 

and  15'  =  lm,  and  V  =4% 

t&'=  Is,  and  I'm  4'. 

Hence  a°  =  4  am,  a!  =  4  a%  a"  =  4a'; 

so  that  to  convert  degrees,  minutes,  fyc.  into  time,  mul- 
tiply by  4,  and  change  the  marks  °  '  "  respectively,  into 


5.  Corollary.  To  convert  time  into  degrees,  minutes, 
fyc,  ?miltiply  the  the  hours  by  15  for  degrees,  and  di- 
vide the  minutes,  seconds,  fyc.  of  time  by  4,  changing 
the  marks  m  s  \  into  °  '  ". 

The  turning  of  degrees,  minutes,  &c.  into  time,  and  the 
reverse,  may  be  at  once  performed  by  table  XXI  of  the  Navi- 
gator. 


6.    Examples. 

1.  Convert  225°  47'  38"  into  time. 

Solution.     By  §  4. 

By 

Table  XXI. 

225°  =  900™  =  15 h 

15* 

47'  =  188s  —           3™  8s 

3m    Qs 

38"=  152'  =                2s  32' 

2*32' 

225°  47'  38"        =    15/l  3™  10'  32' 

15 h  3-  10*32' 

§  7.]  DEFINITIONS.  131 

Arcs  expressed  in  time. 

2.  Convert  17*  19™  13s  into  degrees,  minutes,  &,c. 
Solution.     By  §  5.  By  Table  XXI. 

17*  =  255°  17*  18™        £=  259° 

19-  13*  =       4°  48'  15"  3-  12s  =  48' 


17*  19™  13*  =  259°  48'  15"  Is  dti  15" 


17*  19™  13s  =  259°  48'  15" 

3.  Convert  12°  34'  56"  into  time. 

Ans.     50™  19s  44*. 

4.  Convert  99°  59'  59"  into  time. 

Ans.     6ft  39™  59s  56'. 

5.  Convert  3/l  2™  12s  into  degrees,  minutes,  &c. 

Ans.     45°  33'. 

6.  Convert  11*  59™  59s  into  degrees,  minutes,  &c. 

Ans.     179°  59'  45". 

7.  When  an  arc  is  given  in  time,  its  log.,  sine,  &c. 
can  be  found  directly  from  table  XXVII,  by  means  of 
the  column  headed  Hour  P.  M.,  in  which  twice  the 
time  is  given,  so  that  the  double  of  the  angle  must  be 
found  in  this  column. 

The  use  of  the  table  of  proportional  parts  for  these  columns 
is  explained  upon  page  35  of  the  Navigator.  When  the  time 
exceeds  6*,  the  difference  between  it  and  12*  or  24*  must  be 
used. 


132  SPHERICAL  TRIGONOMETRY.         [CH. 


Arcs  expressed  in  time. 


Examples. 

1.  Find  the  log.  cosine  of  19*  33-  IV. 
Solution. 

24*  —  19*  2Sm  IV  ==  4*  26-  49* 

2  X  (4*  26-  49s)  as  8*  53-  38* 
8*  53"  36s  P.  M.         cos.        9.59720 

prop,  parts  of  2*  7 


8*  53-  38s  P.  M.         cos.         9.59713 

2.    Find   the   angle   in   time  of  which   the   log.   tang,    is 
10.12049. 

T  2-  40s  P.M.     tang.  10.12026 
7*  prop,  parts  23 


2) T  2-  47s  P.M.  10.12049 


Ans.     3*  31-  23J* 

3.  Find  the  log.  sine  of  3*  12-  2s.  Ans.  9.87113. 

4.  Find  the  log.  cosine  of  IP  3-  13s.  Ans.  9.98653. 

5.  Find  the  log.  tang,  of  15*  0m  9*.  Ans.  10.00057. 

6.  Find  the  log.  cotan.  of  22*  59- 59s.  Ans.  10.57183. 

7.  Find  the  angle  in  time  whose  log.  secant  is  10.23456. 

Ans.  3*  37-  26*. 

8.  Find  the  angle  in  time  whose  log.  cosecant  is  10.12346* 

Ans.   3*  15-  15*. 


§  9.]  DEFINITIONS.  133 

Right  and  oblique  triangles. 

8.  An  isosceles  spherical  triangle  is  one,  which  has 
two  of  its  sides  equal. 

An  equilateral  spherical  triangle  is  one,  which  has 
all  its  sides  equal. 

9.  A  spherical  right  triangle  is  one,  which  has  a  right 
angle  ;  all  other  spherical  triangles  are  called  oblique. 

We  shall  in  spherical  trigonometry,  as  we   did  in  plane 
trigonometry,  attend  first  to  the  solution  of  right  triangles. 


id 


134  SPHERICAL    TRIGONOMETRY.  [CH.  II. 

Investigation  of  Neper's  Rules. 


CHAPTER   II. 

SPHERICAL    RIGHT    TRIANGLES. 

10.  Problem.  To  investigate  some  relations  between 
the  sides  and  angles  of  a  spherical  right  triangle. 

Solution.  The  importance  of  this  problem  is  obvious ;  for, 
unless  some  relations  were  known  between  the  sides  and  the 
angles,  they  could  not  be  determined  from  each  other,  and 
there  could  be  no  such  thing  as  the  solution  of  a  spherical 
triangle. 

Let,  then,  ABC  (fig.  29.)  be  a  spherical  right  triangle, 
right-angled  at  C.  Call  the  hypothenuse  AB,  h ;  and  call  the 
legs  BC  and  AC,  opposite  the  angles  A  and  B,  respectively 
a  and  6. 

Let  O  be  the  centre  of  the  sphere.     Join  OA,  OB,  OC. 

The  angle  A  is,  by  art.  2,  equal  to  the  angle  of  the  planes 
BOA  and  CO  A.  The  angle  B  is  equal  to  the  angle  of  the 
planes  BOC  and  BOA.  The  angle  of  the  planes  BOC  and 
AOC  is  equal  to  the  angle  C,  that  is,  to  a  right  angle;  these 
two  planes  are,  therefore,  perpendicular  to  each  other. 

Moreover,  the  angle  BOA,  measured  by  BA,  is  equal  to 
BA  or  A;  BOC  is  equal  to  its  measure  BC  ox  a,  and  AOC 
is  equal  to  its  measure  ^4 Cor  b. 

Through  any  point  A'  of  the  line  OA,  suppose  a  plane  to 
pass  perpendicular  to  OA.     Its  intersections  A'C  and  A'B' 


§   10.]  SPHERICAL    RIGHT    TRIANGLES.  135 

Investigation  of  Neper's  Rules. 

with  the  planes  CO  A  and  BOA  must  be  perpendicular  to 
OA'y  because  they  are  drawn  through  the  foot  of  this  perpen- 
dicular. 

As  the  plane  B'A'C  is  perpendicular  to  OA,  it  must  be 
perpendicular  to  AOC  \  and  its  intersection  B'C  with  the 
plane  BOC,  which  is  also  perpendicular  to  AOC,  must  like- 
wise be  perpendicular  to  AOC.  Hence  B'C  must  be  per- 
pendicular to  A'C  and  OC,  which  pass  through  its  foot  in 
the  plane  AOC. 

All  the  triangles  A' OB',  AOC,  BOC,  and  ABC  are 

then  right-angled ;  and  the  comparison  of  them  leads  to  the 
desired  equations,  as  follows  : 

First.    We  have  from  triangle  A' OB'  by  (4) 

OA1 

cos.  A1  OB'  z=z  cos.  h  —  ; 

and  from  triangles  AOC  and  BOC 

OA' 
cos.  A'OC  ==  cos.  b  =  yr-^-, 


cos.  B'OC  =  cos.  a  = 


oc> 
oc 


OBr 

The  product  of  the  two  last  equations  is 

OA!        OC        OA! 

cos.  a  cos.  b^  —  x-^^^r; 

hence,  from  the  equality  of  the  second  members  of  these  equa- 
tions, 

cos.  h  =z  cos.  a  cos.  b.  (230) 


136  SPHERICAL    TRIGONOMETRY.  [CH.  II. 

Investigation  of  Neper's  Rules. 

Secondly.    From  triangle  A'B  C  we  have  by  (4),  and  the 

fact  that  the  angle  B'A'C  is  equal  to  the  inclination  of  the 

two  planes  BOC  and  BOA, 

A'C 
cos.  B'A'C  =  cos.  A  =z  — —  ; 
A'B1' 

and,  from  triangles  A'OC  and  A' OB',  by  (4), 
tang.  CO  A'  =  tang.  6=  — , 

cotan.  B'OA'  =  cotan.  ^  =  -rrrrr 

A'B' 

The  product  of  these  equations  is 

_       A'C  v    A'O       A'C 
tang,  b  cotan.  A  =  _  X  ^  =  ^; 

hence  cos.  A  —  tang.  6  cotan.  h.  (231) 

Thirdly.  Corresponding  to  the  preceding  equation  between 
the  hypothenuse  h,  the  angle  A,  and  the  adjacent  side  b,  there 
must  be  a  precisely  similar  equation  between  the  hypothenuse 
h,  the  angle  B,  and  the  adjacent  side  a ;  which  is 

cos.  B  z=z  tang,  a  cotan.  h.  (232) 

Fourthly.    From  triangles  BOC,  BOA',   and  B'A'C, 

by  (4), 

B'C 


sin.  ^OC'  =  sin.  a 
sin.  B'OA'  =  sin.  A  = 
sin.  B'A'C  M  sin.  4  = 


OB1' 

B'A' 
~OB'9 

B'C 


B'A' 


§  10.]  SPHERICAL    RIGHT    TRIANGLES.  137 

Investigation  of  Neper's  Rules. 

The  product  of  these  last  two  equations  is 

.      _        BA1       B'C       B'C 

sin.  h  sin.  ^^X^  =  M/J 

hence  sin.  a  5=3  sin.  h  sin.  A.  (233) 

Fifthly.  The  preceding  equation  between  h,  the  angle  A, 
and  the  opposite  side  a,  leads  to  the  following  corresponding 
one  between  h,  the  angle  B,  and  the  opposite  side  b  ; 

sin.  b  ±z  sin.  h  sin.  B.  (234) 


Sixthly.    From  triangles  C'OA'9  BAC,  and  BOC,  by 

(4), 


sin.  CO  A1  =.  sin.  6  == 


cotan.  B'A'C  =  cotan.  -4.  = 


'//^/> 


B'C 

B'C 
tang.  B'OC  ==  tang,  a  =7777/- 

The  product  of  these  last  two  equations  is 

4'C    ,  B'C       A'C 
cotan.  ^  tang,  a  =  —  X  ^  =  ^; 

hence  sin.  6  =  cotan.  J.  tang.  a.  (235) 

Seventhly.  The  preceding  equation  between  the  angle  Ay 
the  opposite  side  a,  and  the  adjacent  side  b,  leads  to  the  fol- 
lowing corresponding  one  between  the  angle  B,  the  opposite 
side  b,  and  the  adjacent  side  a ; 

sin.  a  =  cotan.  5  tang.  b.  (236) 

12* 


138  SPHERICAL    TRIGONOMETRY.  [CIJ.  II. 

Investigation  of  Neper's  Rules. 

Eighthly.    From  (7) 

sin.  a 

tang,  a  — , 

cos.  a 

sin.  b 

tanp*.  b  =. -\ 

&  cos.  6 

which,  substituted  in  (235)  and  (236),  give 

cotan.  B  sin.  b 


sin.  a  s= 


sin.  b  : 


cos.  b 
cotan.  A  sin.  a 


cos.  a 

Multiplying  the  first  of  these  equations  by  cos.  b  and  the 
second  by  cos.  a,  we  have 

sin.  a  cos.  b  =  cotan.  B  sin.  6, 

sin.  b  cos.  a  z=  cotan  .4  sin.  a. 

The  product  of  these  equations  is 

sin.  a  sin.  b  cos.  a  cos.  6  ss  cotan.  ^4  cotan.  2?  sin.  a  sin.  6 ; 

which,  divided  by  sin.  a  sin.  b,  becomes 

cos.  a  cos.  b  sg  cotan.  ^4  cotan.  I?. 

But,  by  (230), 

cos.  h  zz:  cos,  a  cos.  b ; 
hence  cos.  h  =  cotan.  ^4  cotan.  A  (237) 

Ninthly.  We  have,  by  (230)  and  (234), 

cos.  h 

COS.  a  :zz -, 

cos.  6 

_       sin.  b 
sin.  B=z— — r, 
sin.  h 


§  11.]  SPHERICAL    RIGHT    TRIANGLES.  139 

Neper's  Rules. 

the  product  of  which  is,  by  (7)  and  (8), 

_        sin.  b  cos.  h         sin.  6   cos.h 

COS.   a  Sin.  B  =  : : =  r.   - 7 

cos.  6  sin.  a        cos. 6   sin.  ft 
r=  tang,  b  cotan.  h. 
But,  by  (231), 

cos.  A  =  tang,  b  cotan.  It ; 
hence  cos.  A  =  cos.  a  sin.  B.  (238) 

Tenthly.  The  preceding  equation  between  the  side  a,  the 
opposite  angle  A,  and  the  adjacent  angle  B,  leads  to  the  fol- 
lowing similar  one  between  the  side  b9  the  opposite  angle  B , 
and  the  adjacent  angle  A ; 

cos.  B  ±s  cos.  6* sin.  ^4.  (239) 

11.  Corollary.  The  ten  equations,  [230-239],  have, 
by  a  most  happy  artifice,  been  reduced  to  two  very 
simple  theorems,  called,  from  their  celebrated  inventor, 
Neper's  Mules. 

In  these  rules,  the  complements  of  the  hypothenuse 
and  the  angles  are  used  instead  of  the  hypothenuse  and 
the  angles  themselves,  and  the  right  angle  is  neglected. 

Of  the  five  parts,  then,  the  legs,  the  complement  of 
the  hypothenuse,  and  the  complements  of  the  angles  ; 
either  part  may  be  called  the  middle  part.  The  two 
parts,  including  the  middle  part  on  each  side,  are  called 
the  adjacent  parts  ;  and  the  other  two  parts  are  called 
the  opposite  parts.     The  two  theorems  are  as  follows. 

I.  The  sine  of  the  middle  part  is  equal  to  the  product 
of  the  tangents  of  the  two  adjacent  parts. 


140  SPHERICAL    TRIGONOMETRY.  [CH.  II. 

Neper's  Rules. 


II.  The  sine  of  the  middle  part  is  equal  to  the  product 
of  the  cosines  of  the  two  opposite  parts.     [B.  p.  436.] 

Proof.  To  demonstrate  the  preceding  rules,  it  is  only  neces- 
sary to  compare  all  the  equations  which  can  be  deduced  from 
them,  with  those  previously  obtained.   [230  -  239.] 

Let  there  be  the  spherical  right  triangle  ABC  (fig.  30.) 
right-angled  at  C. 

First.  If  co.  h  were  made  the  middle  part,  then,  by  the 
above  rule,  co.  A  and  co.  B  would  be  adjacent  parts,  and  a 
and  b  opposite  parts  ;  and  we  should  have 

sin.  (co.  h)  =  tang.  (co.  A)  tang.  (co.  B) 

sin.  (co.  h)  =  cos.  a  cos.  b ; 
or  cos.  h  =  cotan.  A  cotan.  J5, 

cos.  h  ±3  cos.  a  cos.  b ; 
which  are  the  same  as  (237)  and  (230). 

Secondly.  If  co.  A  were  made  the  middle  part ;  then  co.  h 
and  b  would  be  adjacent  parts,  and  co.  B  and  a  opposite  parts ; 
and  we  should  have 

sin.  (co.  A)  =  tang.  (co.  h)  tang,  b, 
sin.  (co.  A)  =s?  cos.  (co.  jB)  cos.  a; 
or  cos.  A  ===  cotan.  h  tang,  h, 

cos.  A  =  sin.  B  cos.  a  ; 
which  are  the  same  as  (231)  and  (238). 

In  like  manner,  if  co.  B  were  made  the  middle  part,  we 
should  have 


§  12.]  SPHERICAL    RIGHT    TRIANGLES.  141 

Sides,  when  acute  or  obtuse. 

cos.  B  =  cotan.  7i  tang.  af 
cos.  B  =  sin.  A  cos.  b ; 
which  are  the  same  as  (232)  and  (239). 

Thirdly.  If  a  were  made  the  middle  part,  then  co.  B  and 
b  would  be  the  adjacent  parts,  and  co.  A  and  co.  h  the  opposite 
parts ;  and  we  should  have 

sin.  a  ==  tang.  (co.  B)  tang  6, 

sin.  a  =  cos.  (co.  ^1)  cos.  (co.  h) ; 

or  sin.  a  =  cotan.  2£  tang.  6, 

sin.  a  =  sin.  A  sin.  7^ ; 

which  are  the  same  as  (236)  and  (233). 

In  like  manner,  if  b  were  made  the  middle  part,  we  should 
have 

sin.  b  =  cotan.  A  tang,  a, 

sin.  b  =  sin.  B  sin.  h ; 

which  are  the  same  as  (235)  and  (234). 

Having  thus  made  each  part  successively  the  middle  part, 
the  ten  equations,  which  we  have  obtained,  must  be  all  the 
equations  included  in  Neper's  Rules;  and  we  perceive  that 
they  are  identical  with  the  ten  equations  [230-239]. 

12.  Theorem.  The  three  sides  of  a  spherical  right 
triangle  are  either  all  less  than  90°  ;  or  else,  one  is  less 
while  the  other  two  are  greater  than  90°,  unless  one  of 
them  is  equal  to  90°,  as  in  §  16. 

Proof.  When  h  is  less  than  90°,  the  first  member  of  (230) 
is  positive ;  and  therefore  the  factors  of  its  second  member 


142  SPHERICAL    TRIGONOMETRY.  [CH.  II. 

Angle  and  opposite  leg  both  acute  or  obtuse. 

must  either  be  both  positive  or  both  negative ;  that  is,  the  two 
legs  a  and  b  must,  by  PL  Trig.  §  61,  be  both  acute  or  both 
obtuse. 

But  when  h  is  obtuse,  the  first  member  of  (230)  is  negative ; 
and  therefore  one  of  the  factors  of  the  second  member  must 
be  positive,  while  the  other  negative  ;  that  is,  of  the  two  legs 
a  and  6,  one  must  be  acute,  while  the  other  is  obtuse. 

13.  Theorem.  The  hypothenuse  differs  less  from  90° 
than  does  either  of  the  legs,  the  case  of  either  side  equal 
to  90°  being  excepted. 

Proof.  The  factors  cos.  a  and  cos.  b  of  the  second  member 
of  the  equation  (230)  are,  by  (4),  fractions  whose  numerators 
are  less  than  their  denominators.  Their  product,  neglecting 
the  signs,  must  then  be  less  than  either  of  them,  as  cos  a  for 
instance,  or 

cos.  h  <  cos.  a ; 

and  therefore,  by  PL  Trig.  §  69  and  70,  h  must  differ  less  from 
90°  than  a  does. 


14.  Theorem.  An  angle  and  its  opposite  leg  in  a 
spherical  right  triangle  must  be  both  acute,  or  both  ob- 
tuse, or,  by  §  16,  both  equal  to  90°. 

Proof  When  A  is  acute,  the  first  member  of  (238)  is  posi- 
tive, and  therefore  the  factor  cos.  a  of  the  second  member, 
being  multiplied  by  the  positive  factor  sin.  JB  must  be  positive ; 
that  is,  a  must  be  acute.  But  if  A  is  obtuse,  the  first  member 
of  (238)  is  negative,  and  therefore  the  factor  cos.  a  of  the 
second  member  must  be  negative  ;  that  is,  a  must  be  obtuse. 


<§>  16.]  SPHERICAL    RIGHT    TRIANGLES.  143 

One  side  equal  to  90°. 

15.  Theorem.  An  angle  differs  less  from  90°  than 
its  opposite  leg,  the  case  of  either  side,  equal  to  90°, 
being  excepted. 

Proof.  Since  the  second  member  of  (238)  is  the  product 
of  the  two  fractions  cos.  a  and  sin.  B,  the  first  member  must 
be  less  than  either  of  them.     Thus,  neglecting  the  sines, 

cos.  A  <  cos.  a ; 

hence  A  differs  less  from  90°  than  a  does. 

16.  Theorem.  When  in  a  spherical  right  triangle 
either  side  is  equal  to  90°,  one  of  the  other  two  sides  is 
also  equal  to  90°  ;  and  each  side  is  equal  to  its  opposite 
angle. 

Proof.  First.  If  either  of  the  legs  is  equal  to  90°,  the  cor- 
responding factor  of  the  second  member  of  (230)  is,  by  (59), 
equal  to  zero  ;  which  gives 

cos.  h  =  0, 

or,  by  (59), 

h=z90°.  .'j 

Again,  if  we  have 

h  =r  90°, 

it  follows,  from  (59)  and  (230),  that 

0  =  cos.  a  cos.  b, 

and  therefore  either  cos.  a  or  cos.  b  must  be  zero ;  that  is, 
either  a  or  b  must  be  equal  to  90°. 


144  SPHERICAL    TRIGONOMETRY.  [CH.  II. 

Sides  equal  to  90°. 

Secondly.  When  either  side  is  equal  to  90°,  it  follows,  from 
the  preceding  proof,  that 

h  5=  90°  ; 

which  substituted  in  (233)  produces,  by  (60), 

sin.  a  =.  sin*  A  ; 

which  gives 

a  =  A ; 

because,  from  §  14,  a  could  not  be  equal  to  the  supplement 

17.  Corollary.  When  both  the  legs  of  a  spherical 
right  triangle  are  equal  to  90°,  all  the  sides  and  angles 
are  equal  to  90°. 

18.  Theorem.  When  two  of  the  angles  of  a  spherical 
triangle  are  equal  to  90°,  the  opposite  sides  are  also 
equal  to  90°. 

Proof.  For  in  this  case,  one  of  the  factors  of  the  second 
member  of  the  equation  (237)  must,  by  (61),  be  equal  to  zero, 
since  either  A  or  B  is  90°  ;  hence 

cos.  h  =  0  ; 

or,  by  (59), 

k  =  90° ; 

and  the  remainder  of  the  proposition  follows  from  §  16. 

19.  Corollary.  When  all  the  angles  of  a  spherical 
right  triangle  are  equal  to  90°,  all  the  sides  are  also 
equal  to  90°. 


§  20.]  SPHERICAL   RIGHT    TRIANGLES.  145 

Limits  of  the  angles. 

20.  Theorem.  The  sum  of  the  angles  of  a  spherical 
right  triangle  is  greater  than  180°,  and  less  than  360°  ; 
and  each  angle  is  less  than  the  sum  of  the  other  two. 

Proof.  I.  It  is  proved  in  Geometry,  that  the  sum  of  the 
angles  of  any  spherical  triangle  is  greater  than  180°. 

II.  It  is  proved  in  Geometry,  that  each  angle  of  any  spheri- 
cal triangle  is  greater  than  the  difference  between  two  right 
angles  and  the  sum  of  the  other  two  angles.  Hence  if  the 
sum  of  the  two  angles  A  and  B  is  greater  than  180°,  we 
have 

90°  >  A  +  B  —  180°, 

or  A  +  B  <  270°, 

or  A  +  B  +  90°  <  360°  ; 

that  is,  the  sum  of  the  three  angles  is  less  than  360°  ;  for  in 
case  the  sum  of  the  angles  A  and  B  is  less  than  180°,  the 
sum  of  the  three  angles  is  obviously  less  than  360°. 

III.  When  the  right  angle  is  greatest  of  the  three  angles, 
we  have 

90°  +  A  +  B  >  180°, 

or  A  +  B  >  90°  ; 

that  is,  the  greater  angle  is  in  this  case  less  than  the  sum  of 
the  other  two. 

But  if  one  of  the  other  angles  A  is  the  greatest  of  the  three 
angles,  we  have,  by  the  proposition  of  Geometry  last  referred  to, 

B  >  90°  +  A  —  180°, 

or  B  >  A  —  90°, 

or  A  <  B  +  90°  ; 

13 


146  SPHERICAL    TRIGONOMETRY.  [CH.  II. 

Hypothenuse  and  an  angle  given. 

so  that  in  every  case  one  angle  is  less  than  the  sum  of  the 
other  two. 

21.  To  solve  a  spherical  right  triangle,  two  parts 
must  be  known  in  addition  to  the  right  angle.  From 
the  two  known  parts,  the  other  three  parts  are  to  be 
determined,  separately,  by  equations  derived  from  Ne- 
per's Rules.  The, two  given  parts,  with  the  one  to  be 
determined  are,  in  each  case,  to  enter  into  the  same 
equation  These  three  parts  are  either  all  adjacent  to 
each  other,  in  which  case  the  middle  one  is  taken  as  the 
middle  part,  and  the  other  two  are,  by  <§>  11,  adjacent 
parts  ;  or  one  is  separated  from  the  other  two,  and  then 
the  part,  which  stands  by  itself,  is  the  middle  part, 
and  the  other  two  are,  by  §  11,  oppotite  parts. 

22.  Problem.  To  solvo  a  spherical  right  triangle, 
when  the  hypothenuse  and  one  of  the  angles  are  given. 

Solution.  Let  A B C  (fig.  30.)  be  the  right  triangle,  right- 
angled  at  C;  and  let  the  sides  be  denoted  as  in  §  10.  Let  h 
and  A  be  given  ;  to  solve  the  triangle. 

First.  To  find  the  other  angle  B.  The  three  parts,  which 
are  to  enter  into  the  same  equation,  are  co.  /*,  co.  A,  and  co.  B; 
and,  by  §  21,  as  they  are  all  adjacent  to  each  other,  co.  h  is 
the  middle  part,  and  co.  A  and  co.  B  are  adjacent  parts. 
Hence,  by  Neper's  Rules, 

sin.  (co.  h)  =z  tang.  (co.  A)  tang.  (co.  B), 

or  cos.  h  ==  cotan.  A  cotan.  B  ; 

and,  by  (6), 


§  24]  SPHERICAL    RIGHT    TRIANGLES.  147 


Hypothenuse  and  an  angle  given. 


_  cos.  h 

cotan.  B  = =  cos.  /*  tang.  A 

cotan.  A 


Secondly.  To  find  the  opposite  leg  a.  The  three  parts  are 
co.  A,  co.  k,  and  a;  of  which,  by  §  21,  a  is  the  middle  part, 
and  co.  h  and  co.  A  are  the  opposite  parts.  Hence,  by  Neper's 
Rules, 

sin.  a  =  cos.  (co.  h)  cos.  (co.  A)y 
or  sin.  a  —  sin.  h  sin.  A. 

Thirdly.  To  find  the  adjacent  leg  b.  The  three  parts  are 
co.  Ay  co.  h,  and  6 ;  of  which  co.  A  is  the  middle  part,  and 
co.  h  and  b  are  the  adjacent  parts.     Hence,  by  Neper's  Rules, 

sin.  (co.  A)  z=z  tang.  (co.  h)  tang.  6, 

or  cos.  A  =z  cotan.  h  tang,  b  ; 

and,  by  (6), 

cos.  J. 

tang.  6  = j-  z=  tang,  h  cos.  -4. 

a  cotan. h  ° 

23.  Scholium.  The  tables  always  give  two  angles,  which  are 
supplements  of  each  other,  corresponding  to  each  sine,  cosine, 
&/C  But  it  is  easy  to  choose  the  proper  angle  for  the  particu- 
lar case,  by  referring  to  §  12  and  14 ;  or  by  having  regard  to 
the  signs  of  the  different  terms  of  the  equation,  as  determined 
by  PL  Trig.  §  61. 

24.  Scholium.  When  h  and  A  are  both  equal  to  90°,  the 
values  of  cotan.  B  and  tang,  b  are  indeterminate  ;  for  the  nu- 
merators and  denominators  of  the  fractional  values  are,  by 
(59)  and  (61),  equal  to  zero  ;  and   in  this  case  there  are  an 


148  SPHERICAL    TRIGONOMETRY*  [CH.  II. 


Hypothenuse  and  an  angle  given. 


infinite  number  of  triangles  which   satisfy  the  given  values  of 
h  and  A. 

The  problem  is  impossible  by  §  18,  if  the  given  value  of  h 
differs  from  90°,  while  that  of  A  is  equal  to  90°. 

25.  Examples. 

1.  Given  in  the  spherical  right  triangle  (fig.  30.),  1i  =  145° 
and  A  =  23°  28' ;  to  solve  the  triangle. 

Solution, 

h,  cos.       9.91336  *n,     sin.  9.75859,       tang.  9.84523  n, 

4,  tang.     9.63761,  sin.  9.60012,       cos.    9.96251 


jB,cotan.    9.55097  n;  asm.  9.35871;  b  tang.  9.80774  n. 
Ans.  B  =  109°  34'  33",  a  =  13°  12'  12",  b  =  147°  17  1 5". 

2.  Given  in  the  spherical  right  triangle,  (fig.  30.),  h  =  32° 
34',  and  A  =*  44°  44' ;  to  solve  the  triangle. 

Ans.     ^  —  50°    8' 21", 

a  ~  22°  15'  43", 

b  ■=.  24°  24'  19". 

26.  Problem.  To  solve  a  spherical  right  triangle, 
when  its  hypothenuse  and  one  of  its  legs  are  given. 

*  The  letter  n  placed  after  a  logarithm  indicates  it  to  be  the  logarithm 
of  a  negative  quantity,  and  it  is  plain  that,  when  the  number  of  such 
logarithms  to  be  added  together  is  even,  the  sum  is  the  logarithm  of  a 
positive  quantity ;  but  if  odd,  the  sum  is  the  logarithm  of  a  negative 
quantity. 


<§>   28.]  SPHERICAL    RIGHT    TRIANGLES.  149 


Hypothenuse  and  a  leg  given. 


Solution,   Let  ABC  (fig.  30.)  be  the  triangle;  h  the  given 
hypothenuse,  and  a  the  given  leg. 

First.  To  find  the  opposite  angle  A ;  a  is  the  middle  part, 
and  co.  A  and  co.  h  are  the  opposite  parts.     Hence 

sin.  a  z=i  cos.  (co.  h)  cos.  (co.  A)  j 

or  sin.  a  at  sin.  h  sin.  A  ; 


and,  by  (6), 


sin.  a 
sin.  A  =  — — T  =  sin.  a  cosec.  h. 
sin.  A 


Secondly.  To  find  the  adjacent  angle  B ;  co.  JB  is  the  mid- 
dle part,  and  co.  h  and  a  are  the  adjacent  parts.     Hence 

sin.  (co.  B)  ==  tang,  a  tang.  (co.  h), 

or  cos.  J5  :=  tang,  a  cotan.  ^. 

Thirdly.  To  find  the  other  leg  b;  co.h  is  the  middle  part, 
and  a  and  b  are  the  opposite  parts.     Hence 

cos.  h  ss  cos.  a  cos.  6  ; 

and,  by  (6), 

cos.  h  _ 

cos.  o  =  =  sec.  a  cos.  h. 

cos.  a 

27.  Scholium.  The  question  is  impossible  by  §  13,  when  the 
given  value  of  the  hypothenuse  differs  more  from  90°  than  that 
of  the  leg. 

28.  Solution.  When  h  and  a  are  both  equal  to  90°,  it  may 
be  shown,  as  in  §  24,  that  the  values  of  B  and  b  are  indeter- 
minate. 

13* 


150  SPHERICAL    TRIGONOMETRY.  [CH.   II. 


A  leg  and  the  opposite  angle  given. 


29.  Example. 

Given  in  the  spherical  right  triangle  (fig.  30.),  az=  141°  \l> 
and  h  z=z  127°  12' ;  to  solve  the  triangle. 

Ans.  A  =  128°  6f  54", 
B  z=  52°  22'  24', 
b  =z    39°    6'  23". 

30.    Problem.     To  solve  a  spherical  right  triangle, 
when  one  of  its  legs  and  the  opposite  angle  are  given. 

Solution.  Let  ABC  (fig.  30.)   be  the  triangle,  a  the  given 
leg,  and  A  the  given  angle. 

First.  To  find  the  hypothenuse  h ;  a  is  the  middle  part,  and 
co.  h  and  co.  A  are  the  opposite  parts.     Hence 

sin.  a  =z  sin.  h  sin.  A ; 

and,  by  (6), 

.        sin.  a  .  ; 

sin.  h  z= z=  sin.  a  cosec.  A. 

sin.  A 

Secondly.    Te  find  the  other  angle  B ;  co.  A  is  the  middle 
part,  and  a  and  co.  B  are  the  opposite  parts.     Hence 

cos.  A  —  cos.  a  sin.  B  ; 

and,  by  (6), 

cos.  4 

sin.  5  zzz z=  sec.  a  cos.  A. 

cos.  a 

Thirdly.    To  find  the  other  leg.  b ;  6  is  the  middle  part, 
and  a  and  co.  A  are  the  adjacent  parts.     Hence 

sin.  b  z=  tang,  a  cotan.  ^4. 


<§>  34.]  SPHERICAL    RIGHT    TRIANGLES.  151 

A  leg  and  the  opposite  angle  given. 

31.  Scholium.  There  are  two  triangles  ABC  and  A'BC 
(fig.  31.)  formed  by  producing  the  sides  AB  and  AC,  to  the 
point  of  meeting  A1,  both  of  which  satisfy  the  conditions  of 
the  problem.  For  the  side  BC  ox  a,  and  the  angle  A,  or  by 
§  2  its  equal  A',  belong  to  both  the  triangles. 

Now  ABA'  and  AC  A'  are  semicircumferences.  Hence  h', 
the  hypothenuse  of  A'BC,  is  the  supplement  of  A;  b1  is  the 
supplement  of  b ;  and  A'BC  is  the  supplement  of  ABC.  One 
set  of  values,  then,  of  the  unknown  quantities,  given  by  the 
tables,  as  in  §  23,  corresponds  to  the  triangle  ABC,  and  the 
other  set  to  ABC. 

32.  Corollary.  When  the  given  values  of  a  and  A  are 
equal,  the  values  of  h,  B,  and  b  become 

sin.  li  —  1,     sin.  B  z=  1,     sin.  b  =.  1  ; 
or,  by  (60), 

h  =  90°,        B  =  90°,        b  ==  90°  ; 
as  in  §  16. 

33.  Corollary.  When  a  and  A  are  equal  to  90°,  the  values 
of  b  and  B  are  indeterminate,  as  in  §  24. 

34.  Scholium.  The  problem  is,  by  §  14,  impossible,  when 
the  given  values  of  the  leg  and  its  opposite  angle  are  such,  that 
one  is  obtuse,  while  the  other  is  acute,  or  that  one  is  equal  to 
90°,  while  the  other  differs  from  90° ;  and,  by  §  15,  it  is  im- 
possible, when  the  given  value  of  the  angle  differs  more  from 
90°  than  that  of  the  leg. 


152  SPHERICAL  TRIGONOMETRY.         [CH.  II. 


A  leg  and  the  opposite  angle  given. 


35.    Example. 

Given  in  the  spherical  right  triangle,  (fig.  30.),  a  =±  35°  44', 
and  A  =  37°  28' ;  to  solve  the  triangle. 


Ans.     h  =  73°  45'  15"  )  (  h  =  106°  14'  45" 

B  —  IT  54'         \    or    {  B  =  102°    6' 
b  c=  69°  50'  24"  j  (  b  =  110°    9'  36". 


36.  Problem.     To  solve  a  spherical  right  triangle, 
when  one  of  its  legs  and  the  adjacent  angle  are  given. 

Solution.    Let  ^4jBC(fig.  30.)  be  the  triangle,  a  the  given 
leg,  and  B  the  given  angle. 

First.  To  find  the  hypothenuse  h ;  co.  B  is  the  middle  part, 
and  co.  h  and  a  are  adjacent  parts.     Hence 

cos.  B  =  tang,  a  cotan.  h ; 

and,  by  (6), 

cos.  B  _ 

cotan.  h  z= =s  cotan.  a  cos.  2*. 

tang,  a 

Secondly.   To  find  the  other  angle  A ;  co.  J.  is  the  middle 
part,  and  co.  B  and  a  are  opposite  parts.     Hence 

cos.  A  =  cos.  a  sin.  J3. 

Thirdly.   To  find  the  other  leg  b ;  a  is  the  middle  part,  and 
co.  B  and  6  are  adjacent  parts.     Hence 

sin.  a  =  tang,  b  cotan.  J5 ; 

and,  by  (6), 

sin.  «  . 

tang.  6  = =;  =  sin.  a  tang.  J5. 

6  cotan.  ii  & 


§38.]  SPHERICAL    RIGHT    TRIANGLES,  153 


The  legs  given. 


37.    Example. 

Given  in  the  spherical  right  triangle,  (fig.  30.),  a=  118°54/, 
and  B  —  12°  19' ;  to  solve  the  triangle. 

Ans.     h  =.  118°  20' 20", 

A  =-    95°  5#    2", 
b  —    10°  49'  17". 

38.  Problem.     To  solve  a  spherieal  right  triangle, 
when  its  two  legs  are  given. 

Solution.    Let  ^.BC  (fig.  30.)  be  the  triangle,  a  and  b  the 
given  legs. 

First.  To  find  the  hypothenuse  h ;  co.  h  is  the  middle  part, 
a  and  b  are  opposite  parts.     Hence 

cos.  h  =z  cos.  a  cos.  b. 

Secondly.  To  find  one  of  the  angles,  as  A ;  b  is  the  middle 
part,  and  co.  A  and  a  are  adjacent  parts.     Hence 

sin.  b  ±b  tang,  a  cotan.  ^4  ; 

and,  by  (6), 

sin. 

cotan.  A  =  =  cotan.  «  sin.  6. 

tang,  a 

In  the  same  way, 

cotan.  B  ==  cotan.  b  sin.  a. 


154  SPHERICAL    TRIGONOMETRY.  [CH.   II. 


The  angles  given. 


39.    Example. 

Given  in  the  spherical  right  triangle,  (fig.  30.),  a  =i  1°,  and 
b  =  100°  ;  to  solve  the  triangle. 

Ans.     h  —  99°  59'  52/r, 

i=    1°    0'  56", 

J5z=90°  11' 24". 

40.  Problem.     To  solve  a  spherical  right  triangle, 
when  the  two  angles  are  given. 

Solution.    Let  ABC  (fig.  30.)  be  the  triangle,  A  and  B  the 
given  angles. 

First.  To  find  the  hypothenuse  h ;  co.  h  is  the  middle  part, 
and  co.  A  and  co.  B  are  adjacent  parts.     Hence 

cos.  h  =  cotan  A  cotan.  B. 

Secondly.    To  find  one  of  the  legs,  as  a ;  co.  A  is  the  mid- 
dle part,  and  co.  B  and  a  are  the  opposite  parts.     Hence 

cos.  A  33:  cos.  a  sin.  B  \ 

and,  by  (6), 

cos.  A 
cos.  a  =  -^ — =r  =  cos.  J.  cosec.  B. 
sin.  X5 

In  the  same  way, 

cos.  b  —  cosec.  A  cos.  Z?. 

41.  Scholium.    The  problem  is,  by  §  20,  impossible,  when 
the  sum  of  the  given  values  of  A  and  B  is  less  than  90°,  or 


$  42.]  SPHERICAL    RIGHT    TRIANGLES.  155 

The  angles  given. 

greater  than  270°,  or  when  their  difference  is  greater  than 
90.° 


.42.    Example. 

Given  in  the  spherical  right  triangle,  (fig.  30.),  A  =  91°  1 1', 
and  B  =  1110  IV,  to  solve  the  triangle. 

[jins.     h  =    89°  32'  28", 

a  =    91°  16'    8", 

b  =  109°  52  EH 

/  o       * 


156  SPHERICAL    TRIGONOMETRY.  [CH.  Ill 

Sines  of  sides  proportional  to  sines  of  opposite  angles. 

CHAPTER  III. 

SPHERICAL    OBLIQUE    TRIANGLES. 

42.  Theorem.  The  sines  of  the  sides  in  any  spherical 
triangle  are  proportional  to  the  sines  of  the  opposite 
angles.    [B.  p.  437.] 

Proof.  Let  ABC  (figs.  32  and  33.)  be  the  given  triangle. 
Denote  by  a,  b,  c,  the  sides  respectively  opposite  to  the  angles 
A,  By  C.  From  either  of  the  vertices  let  fall  the  perpendicu- 
lar BP  upon  the  opposite  side  AC.  Then,  in  the  right  triangle 
ABPy  making  BP  the  middle  part,  co.  c  and  co.  BAP  are 
the  opposite  parts.     Hence,  by  Neper's  Rules, 

sin.  BP  =±  sin.  c  sin.  BAP  ==  sin.  c  sin.  A. 

For  BAP  is  either  the  same  as  A,  or  it  is  its  supplement,  and 
in  either  case  has  the  same  sine,  by  (91). 

Again,  in  triangle  BPC,  making  BP  the  middle  part,  co.  a 
and  co.  C  are  the  opposite  parts.     Hence,  by  Neper's  Rules, 

sin.  BP  =s=  sin.  a  sin.  C; 

and,  from  the  two  preceding  equations, 

sin.  c  sin.  A  =  sin.  a  sin.  C, 

which  may  be  written  as  a  proportion,  as  follows ; 

sin.  a  :  sin.  A  =  sin  c  :  sin.  C. 

In  the  same  way, 

sin.  a  :  sin.  A  =  sin.  b  .  sin.  B. 


§  43.]  SPHERICAL    OBLIQ.UE    TRIANGLES.  157 

Bowditch's  Rules. 

43.  Theorem.  BowditcK's  Rules  for  Oblique  Tri- 
angles. If,  in  a  spherical  triangle,  two  right  triangles 
are  formed  by  a  perpendicular  let  fall  from  one  of  its 
verticles  upon  the  opposite  side  ;  and  if,  in  the  two 
right  triangles,  the  middle  parts  are  so  taken  that  the 
perpendicular  is  an  adjacent  part  in  both  of  them ;  then 

The  sines  of  the  middle  parts  in  the  two  triangles  are 
proportional  to  the  tangents  of  the  adjacent  parts. 

But,  if  the  perpendicular  is  an  opposite  part  in  both 
the  triangles,  then 

The  sines  of  the  middle  parts  are  proportional  to  the 
cosines  of  the  opposite  parts.    [B.  p.  437.] 

Proof.  Let  M  denote  the  middle  part  in  one  of  the  right 
triangles,  A  an  adjacent  part,  and  O  an  opposite  part.  Also 
let  m  denote  the  middle  part  in  the  other  right  triangle,  a  an 
adjacent  part,  and  o  an  opposite  part ;  and  let  p  denote  the 
perpendicular. 

First.  If  the  perpendicular  is  an  adjacent  part  in  both  tri- 
angles, we  have,  by  Neper's  Rules, 

sin.  M  ==;  tang.  A  tang,  p, 

sin.  m  —  tang,  a  tang,  p  ; 
whence 

sin.  M tang.  A  tang,  p       tang.  A 

sin.  m       tang,  a  tang,  p  ~~  tang,  a  * 

or  sin.  M :  sin.  m  =z  tang.  A  :  tang.  a. 

Secondly.  If  the  perpendicular  is  an  opposite  part  in  both 
the  triangles,  we  have,  by  Neper's  Rules, 

14 


whence 


158  SPHERICAL    TRIGONOMETRY.  [CH.  III. 

Two  sides  and  the  included  angle  given. 

sin.  M  z=z  cos.  O  cos.  pt 
sin.  m  z=  cos.  o  cos.  p  ; 

sin.  31 cos.  O  cos.  p       cos.  O 

sin.  m        cos.  o  cos.  p        cos.  o  ' 

sin.  M :  sin.  m  —  cos.  O  :  cos  o. 


44.  Problem.  To  solve  a  spherical  triangle,  when 
two  of  its  sides  and,  the  included  angle  are  given. 
[B.  p.  438.] 

Solution.  Let  ABC  (figs.  32  and  33.)  be  the  triangle, 
a  and  b  the  given  sides,  and  C  the  given  angle.  From  B  let 
fall  on  .AC  the  perpendicular  BP. 

First.  To  find  PC,  we  know,  in  the  right  triangle  BI£C, 
the  hypothenuse  a  and  the  angle  C.  Hence,  by  means  of 
Neper's  Rules, 

tang.  PC  =  cos.  C  tang.  a.  (239) 

Secondly.  AP  is  the  difference  between  AC  and  PC, 
that  is, 

(fig.  32.)  AP  —  b  —  PC  ot  (fig  33.)  AP  =  PC—  b.  (240) 

Thirdly.  To  find  the  side  c.  If,  in  the  triangle  BPCf 
co.  a  is  the  middle  part,  PC  and  PjB  are  opposite  parts ;  and 
if,  in  the  triangle  ABPt  co.  c  is  the  middle  part,  BP  and  AP 
are  the  opposite  parts.     Hence,  by  Bowditch's  Rules, 

cos.  PC  :  cos.  AP  =  sin.  (co.  a)  :  sin.  (co.  c), 

or  cos.  PC  :  cos.  -4P  z=  cos.  a  :  cos.  c.  (241) 


<§>  45.]  SPHERICAL    OBLIQUE    TRIANGLES.  159 

Rules  for  acute  or  obtuse  angles  and  sides. 


Fourthly,  To  find  the  angle  A.  If,  in  the  triangle  BPC, 
PC  is  the  middle  part,  co.  C  and  BP  are  adjacent  parts;  and 
if,  in  the  triangle  ABP,  AP  is  the  middle  part,  co.  BAP 
and  BP  are  adjacent  parts.     Hence,  by  Bowditch's  Rules, 

sin.  PC  :  sin.  PA  =  cotan.  C  :  cotan.  BAP  ;       (242) 

and  BAP  is  the  angle  A  (fig.  32.),  when  the  perpendicular 
falls  within  the  triangle  ;  or  it  is  the  supplement  of  A  (fig.  33.), 
when  the  perpendicular  falls  without  the  triangle. 

Fifthly.    B  is  found  by  means  of  (344), 

sin.  c  :  sin.  C  =  sin.  b  :  sin.  B.  (243) 

45.  Scholium.  In  determining  PC,  c,  and  BAP,  by  (239), 
(241),  and  (242),  the  signs  of  the  several  terms  must  be  care- 
fully attended  to;  by  means  of  PI.  Trig.  §  61. 

But  to  determine  which  value  of  B,  determined  by  (243), 
is  the  true  value,  regard  must  be  had  to  the  following  rules, 
which  are  proved  in  Geometry. 

I.  The  greater  side  of  a  spherical  triangle  is  always 
opposite  to  the  greater  angle. 

II.  Each  side  is  less  than  the  sum  of  the  other  two. 

III.  The  sum  of  the  sides  is  less  than  360°. 

IV.  Each  angle  is  greater  than  the  difference  between 
180°,  and  the  sum  of  the  other  two  angles. 

There  are,  however,  cases  in  which  these  conditions  are  all 
satisfied  by  each  of  the  values  of  B.  In  any  such  case  this 
angle  can  be  determined  in  the  same  way  in  which  the  angle 
A  was  determined,   by  letting  fall  a  perpendicular  from  the 


160  SPHERICAL  TRIGONOMETRY.        [CH.  III. 

Fundamental  equation. 

vertex  A  on  the  side  BC.  But  this  difficulty  can  always  be 
avoided,  by  letting  fall  the  perpendicular  upon  that  of  the  two 
given  sides  which  differs  the  most  from  90°. 

46.  Corollary.  By  (240),  (104),  and  (29),  we  have 

cos.  AP  =  cos.  (b  —  PC)  s=  cos.  (PC—  b) 

=  cos.  b  cos.  PC  +  sin.  b  sin.  PC,         (244) 

which,  substituted  in  (241),  gives 

cos.  PC:  cos.  b  cos.  PC-\-  sin.  b  sin.  PC=z  cos.  a  :  cos.  c. 

Dividing  the  two  terms  of  the  first  ratio  by  cos.  PC,  we  have 
by  (7), 

1  :  cos.  b  -|-  sin.  b  tang.  PC  r=  cos.  a  :  cos.  c.       (245) 

The  product  of  the  means  being  equal  to  that  of  the  extremes, 
we  have 

cos.  c  z=z  cos.  a  cos.  b  -|-  sin.  b  cos.  a  tang.  PC.        (246) 
But  by  (239) 

tans.  PC  =  cos.  C  tang,  a  = , 

s  5  cos.  a      ' 

or  cos.  a  tang.  PC  =  cos.  C  sin.  a ;  (247) 

which,  substituted  in  (246),  gives 

cos.  c  —=  cos.  a  cos.  6  -f-  sm-  a  sm«  ^  cos«  C>         (248) 

which  is  owe  o/*  the  fundamental  equations  of  Spherical 
Trigonometry. 

47.  Corollary.    We  have,  by  (48), 

cos.  C=zl-+2(cos.  £C)2, 


§    49.]  SPHERICAL    OBLIQ.UE    TRIANGLES.  161 


Column  of  Log.  Rising  of  Table  XXIII. 


which,  substituted  in  (248),  gives,  by  (28), 

cos.  c  ==  cos.  (a  -f-  b)  +  2  sin.  a  sin.  b  (cos.  £  C)2,      (249) 

from  which  the  value  of  the  side  c  can  readily  be   found  by 
using  the  table  of  Natural  Sines. 

48.  Corollary.    We  have,  by  (49), 

cos.  C—  1  —  2  (sin.JC)2, 
which,  substituted  in  (248),  gives,  by  (29), 

cos.  c  —  cos.  (a  —  b)  — 2  sin.  a  sin.  b  (sin.  J  C)2,     (250) 
which  can  be  used  like  formula  (249). 

49.  Corollary.  The  use  of  formula  (250)  is  much  facilitated 
by  means  of  the  column  of  Rising  in  Table  XXIII  of  the 
Navigator.     This  column  contains  the  values  of 

log.  2  (sin.  £  C)2  —  2  log.  sin.  £  C  +  log.  2 

=  2 log.  sin.  iC+  0.30103.    (251) 

But  the  decimal  point  is  supposed  to  be  changed  so  as  to 
correspond  to  the  table  of  Natural  Sines,  that  is,  5  is  added  to 
the  logarithm ;  and  20  is  to  be  subtracted  from  the  value  of 
2  log.  sin.  J  C,  which  is  given  by  table  XXVII,  as  is  evident 
from  PI.  Trig.  §  30.  So  that  the  coltimn  Rising  of  Table 
XXIII  is  constructed  by  the  formula 

log.  Ris.  C  =  2  log.  sin.  J  C  +  5.30102  —  20 

—  2  log.  sin.  J  C  —  14.69897,  (252) 

which    agrees    with  the  explanation  in   the    Preface    to    the 
Navigator. 

14* 


162  SPHERICAL    TRIGONOMETRY.  [CH.  III. 

Two  sides  and  the  included  angle  given. 


50.  By  using  table  XXIII,  the  following  rule  is  ob- 
tained for  finding  the  third  side,  when  two  sides  and 
the  included  angle  are  given. 

Add  together  the  log.  Rising  of  the  given  angle,  and 
the  log.  msines  of  the  two  given  sides.  The  sum  is  the 
logarithm  of  a  number,  which  is  to  be  subtracted  from 
the  natural  cosine  of  the  difference  of  tioo  given  sides 
{regard  being  had  to  the  sign  of  this  cosine).  The  dif- 
ference is  the  natural  cosine  of  the  required  side. 

51.     Examples. 


1. 
Si 

Calculate  the  value  of  log. 
)lution> 

£  (4*28w)  =  2h  Um 

log.  Ris.  4*28" 

Ris. 

of  4A 
sin 

28-. 

9.74189 
2 

19.48378 
14.69897 

4.78481 

2.  Given  in  the  spherical  triangle  two  sides  equal  to  45°  54', 

and  138°  32,  and  the  included   angle  98°  44';  to  solve  the 
triangle. 

Solution.    I.  by  (239), 

€=:    98°  44/  cos.  9.18137  n. 

a  ==    45°  54'  tang.  0.01365 

PC  =171°    6'  16"        tang.  9.19502  n. 


§  51.]  SPHERICAL    OBLIQUE    TRIANGLES.  163 

Two  sides  and  the  included  angle  given.       « 

By  (940), 

AP  =  171°    6'  16"  —  138°  32'  ==  32°  34'  16". 

By  (241), 

PC  =  171°    6'  16"  cos.  (ar.co.)  10.00525  n. 

AP  =    32°  34'  16"  cos.                    9.92569 

a  =    45°  54'  cos.                    9.84255 


c  =  126°  24'  45"  cos.  9.77349  w. 

By  (242), 

PC  =  171°    6'  16"  sin.  (ar.  co.)  10.81071 

AP  =    32°  34'  16"  sin.  9.73106 

C  =    98°  44'  cotan.  9.18644  ». 


BAP  =  118°    8'  19"        cotan.  9.72821  n. 

A  =  180°  —  118°  8'  19"  =3  61°  51'  41". 

By  (243), 

c  =  126°  24'  45"  sin.  (ar.  co.)  10.09433 

C  =    90°  44'  sin.  9.99494 

b  =  138°  32'  sin.  9.82098 


B  =  125°  34' 48"'      sin.  9.91025 

Ans.  c  =  126°  24'  45" 
A  =  61°  61'  41" 
B  =  125°  34'  48". 


164  SPHERICAL    TRIGONOMETRY.  [CH.   III. 

•       Two  sides  and  the  included  angle  given. 

II.  The  third  side  is  thus  caloulated  by  means  of  (249), 

2  log.     0.30103 

45°  54'  sin.     9.85620 

138°  32'  sin.     9.82098 

£(98°  44')  =    49°  22'  2  cos.  19.62744 


0.40332  9.60565 

—  0.99683  =  Nat.  cos.  (13S°  32'  +  45°  54')=  N.  cos.  18426 


_  0.59351  =  Nat.  cos  126°  24'  23"  =  c. 

III.  The  third  side  is  thus  calculated  by  §  41. 

98°  44'  «  6*  34™  56'       '  log.  Ris.  5.06139 

45°  54'  sin.  9.85620 

138°  32'  sin.  9.82098 


54774  4.73857 

92°  38'        N.  cos.  —    4594 


c  =  126°  25'  8"     N.  cos.  —  59368 

3.  Calculate  the  log.  Ris.  of  HM2m  20*. 

Ans.     5.29632. 

4.  Given  in  a  spherical  triangle  two  sides  equal  to  100°,  and 
125°,  and  the  included  angle  equal  to  45° ;  to  solve  the  tri- 
angle. 

Ans.     The  third  side  =  47°  55'  52" 

The  other  two  angles  —  69°  43'  48",  and  =  128°  42' 48". 


§  52.]  SPHERICAL    OBLIQUE    TRIANGLES.  165 


A  side  and  the  two  adjacent  angles  given. 


52.  Problem.  To  solve  a  spherical  triangle,  when 
one  of  its  sides  and  the  two  adjacent  angles  are  given. 
[B.  p.  438.] 

Solution.  Let  ABC  (figs.  32  and  33.)  be  the  triangle  ;  a 
the  given  side,  and  B  and  C  the  given  angles.  From  B  let 
fall  on  AC  the  perpendicular  BP. 

Fir$t.  To  find  PBCf  we  know,  in  the  right  triangle  BPCy 
the  hypothenuse  a  and  the  angle  C.  Hence,  by  Neper's 
Rules, 

cotan.  PBC  =  cos.  a  tang.  C.  (253) 

Secondly.  ABP  is  the  difference  between  ABC  and  PBC, 
that  is, 

(fig.  32.)  ABP  =B  —  PBC, 
or  (fig.  33.)  ABP  =  PBC  —  B.  (254) 

Thirdly.  To  find  the  angle  A.  If,  in  the  triangle  PBC, 
co.  C  is  the  middle  part,  PB  and  co.  PBC  are  the  opposite 
parts;  and  if,  in  the  triangle  ABP,  co.  BAP  is  the  middle 
part,  PB  and  co.  ABP  are  the  opposite  parts.  Hence,  by 
Bowditch's  Rules, 
cos.  (co.  PBC) :  cos.  (co.  ABP)  —  sin.  (co.  C) :  sm.(co.BAP), 

or  sin.  PBC :  sin.  ABP  =  cos.  C :  cos.  BAP  ;  (255) 
and  BAP  is  either  the  angle  A  or  its  supplement. 

Fourthly.  To  find  the  side  c.  If,  in  the  triangle  PBC, 
co  PBC  is  the  middle  part,  PB  and  co.  a  are  the  adjacent 
parts ;  and  if,  in  the  triangle  ABP,  co.  ABP  is  the  middle 
part,  PB  and  co.  c  are  the  adjacent  parts.  Hence,  by  Bow- 
ditch's  Rules, 

cos.  PBC  :  cos.  ABP  s=  cotan.  a  :  cotan.  c.         (256) 


166  SPHERICAL    TRIGONOMETRY.  [CH.  III. 


A  side  and  the  two  adjacent  angles  given. 


Fifthly,    b  is  found  by  the  proportion 

sin.  C  :  sin.  c  s=  sin.  B  :  sin.  b.  (257) 

53.  Scholium.  In  determining  PBC,  BAP,  and  c  by  (253), 
(255),  and  (256),  the  signs  of  the  several  terms  must  be  care- 
fully attended  to,  by  means  of  PI.  Trig.  §  61. 

To  determine  which  value  of  b,  obtained  from  (257),  is  the 
true  value,  regard  must  be  had  to  the  rules  of  §  45.  #But  if 
all  these  conditions  are  satisfied  by  both  values  of  b,  then  b 
may  be  calculated  by  letting  fall  a  perpendicular  from  C  on 
the  side  c  in  the  same  way  in  which  c  has  been  obtained  in 
the  preceding  solution.  But  this  case  can  be  avoided  by  let- 
ting fall  the  perpendicular  from  the  vertex  of  that  one  of  the 
two  given  angles,  which  differs  the  most  from  90°. 

54.  Corollary.  Since  180°  —  a,  180°  —  b,  and  180°  —  c 
are  the  angles  of  the  polar  triangle,  and  180°  —  A,  180°  —  B, 
and  180°  —  C  are  its  sides ;  we  have  given  in  the  polar  tri- 
angle the  two  sides  180°  — jB,  and  180°  — -  C,  and  the  in- 
cluded 180°  —  a ;  so  that  the  polar  triangle  might  be  solved 
by  §  44. 

55.  Corollary,  If  formula  (248)  is  applied  to  the  polar  tri- 
angle of  the  preceding  section,  it  becomes  by  PI.  Trig.  §  60, 

—  cos.  A  =  cos.  B  cos.  C —  sin.  B  sin.  C cos.  a, 

or  cos.  A*—  —  cos.  jB  cos.  C+sin.jBsin.  Ccos.  a.    (258) 

56.  Corollary.  In  the  same  way  (249)  becomes  by  (92) 
and  (116), 

cos.  A  =  —  cos.  (B  +  C )  —2  sin.  B  sin.  C(sin.  £  a)*,    (259) 


§58.] 


SPHERICAL    OBLIQUE    TRIANGLES. 


167 


A  side  and  the  two  adjacent  angles  given. 


from   which  the  value  of  the  third   angle  may  be  found  by 
means  of  table  XXIII. 

57.  Corollary.  In  the  same  way  (250)  becomes  by  (92), 
cos.  A=z  — cos.(J3  —  C)+2sin.jBsin.C(cos.£a)2,    (260) 
from  which  the  value  of  the  third  side  may  be  found. 


58.    Examples. 

1.  Given  in  a  spherical  triangle  one  side  equal  to  175°  27', 
and  the  two  adjacent  angles  equal  to  126°  12',  and  109°  16'; 
to  solve  the  triangle. 


Solution.    I.  By  (253), 

a  =  175°  27' 

COS. 

9.99863  n. 

C  =  109°  16' 

tang. 

0.45650  n. 

PBC  =    19°  19'  24"        cotan. 


0.45513 


By  (254), 

ABP  =  126°  12'  —  19°  19'  24"  =  106°  52  36". 


By  (255), 

PBC—    19°  19' 24" 

ABP  =  106°  52'  36" 

C  =  109°  16' 

BAP  =  162°  36' 


sin.  (ar.  co.)  10.48031 
sin.  9.98088 

cos.  9.51847  n. 


COS. 


9.97966  n. 


168  SPHERICAL    TRIGONOMETRY.  [CH.  III. 


A  side  and  the  two 

adjacent  angles  given. 

By  (256), 

PBC  =    19°  19'  24" 

cos.  (ar.  co.)  10.02518 

ABP  ===  106°  52'  36" 

cos.                  9.46288  n. 

a  =  175°  27' 

cotan.              1.09920  n. 

c  =    14°  30'    9" 

cotan.              0.58726 

.4  =  .R4P 

=  162°  36'. 

• 

By  (257), 

• 

C  =  109°  16' 

sin.  (ar.  co.)  10.02503 

c  =    14°  30'    9" 

sin.                    9.39867 

B  =  126°  12' 

sin.                   9.90685 

b  =  167°  38'  21" 

sin.                    9.33055 

Ans.     A  =  162°  36' 

b  =  167°  38'  21" 

cz=z    14°  30'    9". 

II.  The  third  angle  is  thus  calculated  by  means  of  (259). 
175°  27'  =  1 1 h  41 m  48*  log.  Ris.  5.30035 

126°  12'  .      sin.  9.90685 

109°  16'  sin.  9.97497 


—  152115  5.18217 

235°  28'        —  N.  cos.  56689 


162°  36  12"      N.cos.    —  95426 


§  59.]  SPHERICAL    OBLIQUE    TRIANGLES.  169 

Two  sides  and  an  opposite  angle  given. 

III.  The  third  angle  is  thus  calculated  by  means  of  (260), 

2  log.     0.30103 

J  (175°  27)  =  87°  43'  30"      2  cos.  17.19750 

126°  12'  sin.  9.90685 

109°  16'  sin.  9.97497 


0.00240       7.38035 
16°  56'  —  N.  cos.  —  0.95664 


A  =  162°  36'    N.  cos.  —  0.95424 

2.  Given  in  a  spherical  triangle  one  side  =  45°  54', 
and  the  two  adjacent  angles  =z  125°  37',  and  =  98°  44' ;  to 
solve  the  triangle. 

Ans.     The  third  angle  =  61°  55'  2", 
The  other  two  sides  =  138°  34'  22",  and  =  126°  26' 11". 

59.  Problem.  To  solve  a  spherical  triangle,  token 
two  sides  and  an  angle  opposite  one  of  them  are  given. 
[B.  p.  437.] 

Solution.  Let  ABC  (figs.  32  and  33.)  be  the  triangle,  a  and 
c  the  given  sides,  and  C  the  given  angle.  From  B  let  fall  on 
AC  the  perpendicular  BP. 

First.  To  find  PC.  We  know,  in  the  right  triangle  PBC, 
the  side  a  and  the  angle  C.     Hence,  by  Neper's  Rules, 

tang.  PC  z=l  cos.  Ctang.  a.  (261) 

Secondly.  To  find  AP.     If,  in  the  triangle  PBC,  co.  a  is 
the  middle  part,  CP  and  PB  are  the  opposite  parts;  and  if, 
15 


170  SPHERICAL    TRIGONOMETRY.  [CH.   III. 

Two  sides  and  an  opposite  angle  given. 

in  the  triangle  ABP,  co.  c  is  the  middle  part,  AP  and  PB 
are  the  opposite  parts.     Hence,  by  Bowditch's  Rules, 

cos.  a  :  cos.  c  —  cos.  PC  :  cos.  AP.  (262) 

Thirdly.  To  find  b.  There  are,  in  general,  two  triangles 
which  resolve  the  problem,  in  one  of  which  (fig.  32.) 

b  =  PC  +  AP,  (263) 

and  in  the  other  (fig.  33.) 

b  =  PC—  AP.  (264) 

But,  if  AP  is  greater  than  PC,  there  is  but  one  triangle, 
as  in  (fig.  32.),  and  b  is  obtained  by  (263) ;  or,  if  the  sum  of 
AP  and  PC  is  greater  than  180°,  there  is  but  one  triangle, 
as  in  (fig.  33.),  and  b  is  obtained  by  (264). 

Fourthly.  A  and  B  are  found  by  the  proportion 

sin.  c  :  sin.  C  =.  sin.  a  :  sin.  A  (265) 

sin.  c  :  sin.  C  =.  sin.  b  :  sin.  B.  (266) 

60.  Scholium.  In  determining  PC  and  AP  by  (261)  and 
(262),  the  signs  of  the  several  terms  must  be  carefully  at- 
tended to  by  means  of  PL  Trig.  §  61. 

The  two  values  of  A,  given  by  (265),  correspond  respective- 
ly to  the  two  triangles  which  satisfy  the  problem  ;  and  the 
one,  which  belongs  to  each  triangle,  is  to  be  selected,  so  that 
the  angle  BAP,  which  is  the  same  as  A  in  (fig.  32.),  and 
the  supplement  of  A  in  (fig.  33.),  may  be  obtuse  if  C  is  ob- 
tuse, and  acute  if  C  is  acute.  For  BP  is  the  side  opposite 
BAP  in  the  right  triangle  ABP,  and  the  side  opposite  C  in 
the  triangle  BCP ;  and  therefore,  by  §  14,  BP,  BAP,  and 
C  are  all  at  the  same  time  acute,  or  all  obtuse. 


§  62.]  SPHERICAL    OBLIQUE    TRIANGLES.  171 

Two  sides  and  an  opposite  angle  given. 

Of  the  two  values  of  B,  given  by  (266),  the  one  which  be- 
longs to  each  triangle  is  to  be  determined  by  means  of  the 
rules  of  §  45. 

61.  Scholium.  The  problem  is,  by  a  proposition  of  Geome- 
try, impossible,  when  the  given  value  of  c  differs  more  from 
90°  than  that  of  a ;  if,  at  the  same  time,  the  value  of  one  of 
the  two  quantities,  c  and  C,  is  acute,  while  that  of  the  other 
is  obtuse.  And  in  this  case  we  should  find  that  AP  was 
larger  than  PC,  and  at  the  same  time  that  the  sum  of  AP 
and  PC  was  more  than  180°. 


62.  Examples. 

1.  Given  in  the  spherical  triangle,  one  side  z=  35°,  a  second 
side  =  142°,  the  angle  opposite  the  second  side  =  176°  ;  to 
solve  the  triangle. 


Solution.     By  (261), 

C  —  176° 

cos. 

9.99894  n. 

a=L    35° 

tang, 
tang. 

9.84523 

PC  =  145°    3'  56" 

9.84417  n. 

By  (262), 

a—    35° 

cos.  (ar. 

co.)  10.08664 

PC  —.  145°    3'  56" 

cos. 

9.91371  n. 

c  ==  142° 

cos. 

9.89653  n. 

AP  =    37°  56'  30"         cos.  9.89688 


172  SPHERICAL    TRIGONOMETRY.  [CH.  III. 


Two  sides  and  an 

opposite  angle  given. 

By  (264), 

b  =  145° 

3'  56"  — 

37°  56'  30"  =  107°  7/  26". 

By  (265), 

c  =  142° 

sin.  (ar.co.)  10.21066 

C  =z  176° 

sin.                  8.84358 

a—    35° 

43'  34" 

sin.                 9.75859 

A=      3° 

sin.                 8.81283 

By  (266), 

c  ±s  142° 

sin.  (ar.co.)  10.21066 

C=  176° 

sin.                  8.84358 

b  ==  107° 

r  26" 

12'  58" 

sin.                 9.98030 

J5=      6° 

sin.                  9.03454 

Am 

r.     b  =  107°    7'  26" 

A  =      3°  43'  34" 

B  =      6°  12'  58". 

2.  Given  in  a  spherical  triangle,  one  side  ==  54°,  a  second 
side  rz=  22°,  the  angle  opposite  the  second  side  =  12°  ;  to 
solve  the  triangle. 

Am.  The  third  side  —    73°  14'  29",  or  =    33°  32'  59". 

One  angle  =    26°  40'  49",  or  =  153°  19'  11". 

The  third  angle  =  147°  53'  51",  or  =     17°  51'  43". 

63.  Problem.  To  solve  a  spherical  triangle,  when 
two  angles  and  a  side  opposite  one  of  them  are  given. 
[B.  p.  438.] 


§  64.]  SPHERICAL    OBLIQ.UE    TRIANGLES.  173 


Two  angles  and  an  opposite  side  given. 


Solution,  Let  ABC  (figs.  32  and  33.)  be  the  triangle,  A 
and  C  the  given  angles,  and  a  the  given  side. 

From  B  let  fall  on  AC  the  perpendicular  BP.  This  per- 
pendicular must  fall  within  the  triangle  if  A  and  C  are  either 
both  obtuse  or  both  acute ;  but  it  falls  without,  if  one  is  obtuse 
and  the  other  acute. 

First   PC  may  be  found  by  (261). 

Secondly,  To  find  AP.  If,  in  the  triangle  PBC,  PC  is 
the  middle  part,  co.  C  and  PB  are  the  adjacent  parts ;  and  if, 
in  the  triangle  ABP,  AP  is  the  middle  part,  co.  BAP  and 
BP  are  the  adjacent  parts.     Hence,  by  Bowditch's  Rules, 

cotan.  C  :  cotan.  BAP  =  sin.  PC  :  sin.  AP.         (267) 

Thirdly.    To  find  b.     We  have 

(fig.  32.)         b  =  PC+  AP,   '  '  (268) 

(fig.  33.)         b  =  PC—  AP.  (269) 

Fourthly,    c  and  B  are  found  by  the  proportion 

sin.  A  :  sin.  a  =  sin.  C :  sin.  c,  (270) 

sin.  a   :  sin.  A  =  sin.  b    :  sin.jB.  (271) 

64.  Scholium.  Either  value  of  AP,  given  by  (267),  may  be 
used,  and  there  will  be  two  different  triangles  solving  the 
problem,  except  when  AP  -\-  PC  (fig.  32.)  is  greater  than 
180°,  or  PC  (fig.  33.)  is  less  than  AP.  It  may  be  that  both 
values  of  AP  satisfy  the  conditions  of  the  problem,  or  that 
only  one  value  satisfies  them,  or  that  neither  value  does ;  in 
which  last  case  the  problem  is  impossible. 
15* 


174  SPHERICAL    TRIGONOMETRY.  [CH.  III. 


Two  angles  and  an  opposite  side  given. 


Of  the  values  of  c,  determined  by  (270),  the  true  value  must 
be  ascertained  from  the  right  triangle  ABP  by§  12;  or  since 
PB  and  C  are  both  acute  or  both  obtuse  at  the  same  time ;  it 
follows,  from  §  12,  that  when  C  and  AP  are  both  acute  or 
both  obtuse,  that  c  is  acute  ;  but  when  one  of  them  is  obtuse 
and  the  other  acute,  c  is  obtuse. 

From  the  two  values  of  B  (271),  the  true  value  must  be 
selected  by  means  of  the  rules  of  §  45. 

65.  Scholium.  The  problem  is  impossible,  by  Geometry, 
when  A  differs  more  from  90°  than  does  C,  and  when  at 
the  same  time  one  of  the  two  quantities  a  and  A  is  acute, 
while  the  other  is  obtuse.  This  case  is  precisely  the  same  as 
the  impossible  case  of  §  61. 

66.  Examples. 
\, 

1.  Given  in  a  spherical  triangle,  one  angle  ±=  95°,  a  second 
angle  —  104°,  and  the  side  opposite  the  first  angle  z=  138°  ; 
to  solve  the  triangle. 
Solution.    By  (261), 

C  =  104°  cos.  9.38368  n. 

a  5=  138°  tang.  9.95444  n. 


PC  z=    12°  17'  20" 

tang.                9.33812 

By  (267), 

C  =  104° 

cotan.  (ar.co.)  0.60323  n, 

PC  =±    12°  17'  20 " 

sin.                    9.32802 

BAP-    95° 

cotan.                8.94195  n, 

AP  s=      4°  16'  59"       sin.  8.87320 


§  66.]  SPHERICAL    OBLIQUE    TRIANGLES.  175 

Two  angles  and  an  opposite  side  given. 

By  (268), 

b  =  12°  17'  20"  +  4°  16'  59'  =  16°  34'  19". 

By  (270), 

A  =    95°  sin.  (ar.co.)  10.00166 

a  =  138°  sin.  9.82551 

C  =  104°  sin.  9.98690 


c  =  139°  19'  40"  sin.                 9.81407 

By  (271), 

a  =  138°  sin.  (ar.co.)  10.17449 

A=    95°  sin.                 9.99834 

b  =    16°  34'  19"  sin.                  9.45518 


B  =    25°    7' 38"        sin.  9.62801 

Again,  by  (269), 

b  =  12°  17'  20"  —  4°  16'  59"  =  7°  0'  21" 
c  =  180°  —  139°  19'  40"  =  40°  40'  20". 

By  (271), 

a  =  138°  sin.  (ar.  co.)  10.17449 

A—    95°  sin.                 9.99834 

b=      7°  021"         sin.                 9.08623 


B=z     10°  27' 42"         sin.  9.25906 

Ans.     b  ss    16°  34'  19"  or  =    7°    0'  21" 

c  =  139°  19'  40"  or  =  40°  40'  20" 

B  =    25°    7'  38"  or  =  10°  27'  42". 


176  SPHERICAL    TRIGONOMETRY.  [CH.   111. 

The  three  sides  given. 

2.  Given  in  a  spherical  triangle,  one  angle  =  104°,  a  second 
angle  ==  95°,  and  the  side  opposite  the  first  angle  =  138°  ;  to 
solve  the  triangle. 

Ans.     The  two  sides  are  17°  22'  13",  and  136°  36' 27". 
The  other  angle  is  25°  39'  9". 

67.  Problem.  To  solve  a  spherical  triangle,  when  its 
three  sides  are  given.  # 

Solution.  Equation  (248)  gives,  by  transposition  and  di- 
vision, 

-,        cos.  c  —  cos.  a  cos.  b  igj*^ 

COS.  C  s=  : : - ,  (272) 

sin.  a  sin.  b  ' 

whence  the  value  of  the  angle  C  may  be  calculated,  and  in 
the  same  way  either  of  the  other  angles. 

68.  Corollary.  An  equation,  more  easy  for  calculation  by 
logarithms,  may  be  obtained  from  (249),  which  gives,  by 
transposition  and  division, 

3  (cos.  $  CY  =  «~ '-<"■;  <«+*).  (273) 

v  '  sm.  a  sin  b  ' 

Now,  letting  s  denote  half  the  sum  of  the  sides,  or 

s  =  i(a  +  b  +  c);  (274) 

if  we  make  in  (35) 

M=i(a  +  b  +  c)=,s, 

N=  %  (a-\-b  —  c)  =  s  —  c; 

we  have 

M  +  N  =  a  +  b, 

M—N=  c; 


§  70.]       SPHERICAL  OBLIQUE  TRIANGLES.         177 

The  three  sides  given. 

and  (35)  becomes 

cos.  c  —  cos.  (a  -j-  b)  =  2  sin.  s  sin.  (s  —  c)  ; 
which,  substituted  in  (273),  gives 

2  (cos.  i  Cy  =  »*»;»«M' --**?;  (275) 

7  sin.  a  sin.  o 


and 

cos 


^n^£&%     ^ 


69.  Corollary.    The  angles  .4  and  JB  may  be  found  by  the 
two  following  equations,  which  are  easily  deduced  from  (276), 

/sin.  5  sin.  (s  —  a)\ 
co,^  =  V(      sm.6sin.T^)'  %& 

\       sin.  a  sin.  c      /  v 


cos, 


70.  Corollary.  Another  equation,  equally  simple  in  calcula- 
tion, can  be  obtained  from  (250),  which  gives,  by  transposition 
and  division, 

cos.  (a  —  6)  —  cos.  a  cos.  b  -j-  sin.  a  sin   6, 

which,  substituted  in  the  numerator  of  (250),  gives 

2  (sin.  j  cy  -  cos.  (.-6) -cos., 
v        2      y  sin.  a  sin.  6  '  v       y 

whence  C  can  be  found  by  Table  XXIII. 

71.  Corollary.    If,  in  (35),  we  make 

M  =  |  (a  —  6  +  c)  =  s  —  6, 
iV=£( —  a  -|-  &  -{-  c)  =  *  —  a, 


178  SPHERICAL    TRIGONOMETRY.  [CH.  III. 

The  three  sides  given. 

we  have 

M  +  N  =  c, 

M—N=  a  —  b, 
and  (35)  becomes 

cos.  (a  —  b)  —  cos.  c  =  2  sin.  (5  —  a)  sin.  (5  —  b)  ; 
which,  substituted  in  (279),  gives 

2(sin.£C)3=  2sin.(s-fl)sin(5-6j 
v        2      '  sin.  a  sin.  6  v       ' 


and 


sin.  |C=V  P'(ff*)Tf^*)fr         (281) 
2  ^    \  sin.  a  sin.  b  ) 


72.  Corollary.   In  the  same  way  we  might  deduce  the  fol- 
lowing equations ; 

sin.  i  A  =  v(^.(s-b)  sin,  (s-c)^ 
2  ™  \  sin.  6  sin.  c  / 

sin.  X  B  =  v/!in_(^-«)sin.(s-C)V 
\  sin.  a  sin.  c  / 

73.  Corollary.    The  quotient  of  (282),  divided  by  (277),  is 
°y  (7), 

tang.  J4*=  S-N4  =  V  /sin-^-6)s;n-(s-i)\,  (284) 
cos.  £  A  \      sm.  5  sm.  (s  —  a)       / 

In  the  same  way, 

n  ./sin.  (s  —  a)  sin.  (s  —  c)\  /nDn 

tang.  A  B  =z  \/( ^ J--, — ^-r — '-  I ,         (285) 

5   2  ^\       sin.  s  sin.  (s  —  b)        /  v       J 


*74] 


SPHERICAL    OBLIQUE    TRIANGLES. 


179 


The  three  sides  given. 


74.    Examples. 

I.    Given  in  the  spherical   triangle  ABC  the  three  sides 
equal  to  46°,  72°,  and  68°  ;  to  solve  the  triangle. 

Solution.    I.  By  (277),  by  (278),  by  (279), 

a=46°sin.  (ar.co.)10.14307(ar.co.)10.14307 

6=72°sin.  (ar.co.)  10.02 179  (ar.co.)10.02179 

c=68°  sin.  (ar.co.)  10.03283  (ar.co.)  10.03283 


5=93°  sin. 

9.99940 

9.99940 

9.99940 

s-a=47°sin. 

9.86413 

s-6=21°sin. 

9.55433 

s-c=25°  sin. 

2) 

9.62595 

19.91815 

2)19.72963 

2)19.79021 

cos.  9.95908  9.86482  9.89511 

}  A  =  24°  29',  £  B  =  42°  54',  $  C  =  38°  14/  18" ; 

Ans.     A  =  48°  58',     B  =  85°  48',      C  =  76°  28'  56". 


II.  By  Table  XXIII  and  equation  (279), 
a  —  b  =  26°     N.  cos.  89879 
c  =  68°     N.  cos.  ,37461 


52418 


a  =  46°     . 

b  =  72° 

C=z  5*5*55*  =  76°28/45// 


log.  4.71948 
sin.  (ar.co.)  0.14307 
sin.  (ar.co.)    0.02179 


log.  Ris.     4.88434 


180  SPHERICAL    TRIGONOMETRY.  [CH.  III. 


Neper's  Analogies. 


2.  Given  in  a  spherical  triangle  the  three  sides  equal  to  3°, 
4°,  and  5° ;  to  solve  the  triangle. 

Ans.     The  three  angles  are  36°  54',  53°  10',  and  90°  2'. 

75.  Neper  obtained  two  theorems  for  the  solution  of 
a  spherical  triangle,  when  a  side  and  the  two  adjacent 
angles  are  given,  by  which  the  two  sides  can  be  calcu- 
lated without  the  necessity  of  calculating  the  third 
angle.  These  theorems,  which  are  given  in  §  78  and 
79,  can  be  obtained  from  equations  (284-286)  by  the 
assistance  of  the  following  lemmas. 

76.  Lemma.    If  we  have  the  equation 

tang.  M       x 
tang.  N        y' 

we  can  deduce  from  it  the  following  equation, 

sin.  (M  +  N)  __  x  +  y 
sin.  (M—  N)  T  x  —  y' 

Proof.    We  have  from  (7) 

sin.  M        .  ,_       sin.  N 

XdLiciv.Mz=. —f  and  tang.  iV  = —  : 

5    -        cos.  if'  s  cos.^V' 

which,  substituted  in  (287),  give 

sin.  M cos.  N       x 
cos.  M  sin.  N       y 

This  equation  is  the  same  as  the  proportion 

sin.  Mcos.  N  :  cos.  if  sin.  N  =  x  :  y  ; 
hence,  by  the  theory  of  proportions, 


(287) 


(288) 


(290) 


<§>  78.]  SPHERICAL    OBLIQUE    TRIANGLES.  181 

Neper's  Analogies. 

sin.  M  cos.  N  +  cos.  M  sin.  N  :  sin.  M  cos.  N 
— -  cos.  M  sin.  iV  =±  x  -\-  y  :  a;  —  y, 

or,  by  (26)  and  (27), 

sin.  {M-\-  N)  :  sin.  (IT—  2V)  z=  x  +  y  :  a;  —  y ; 
which  may  be  written  in  the  form  of  an  equation,  as  in  (288). 

77.  Lemma.  If  we  have  the  equation 

tang.  i*f  tang.  iV  =  -;  (289) 

we  can  deduce  from  it  the  equation 

cos.  ( M  -f-  N)  __  y  —  x 
cos.  (M  —  N)  *~"  y-\-  x 

Proof.    We  have,  by  (289)  and  (7), 

sin.  M  sin.  N x 

cos.  M  cos.  N       y 
This  equation  is  the  same  as  the  proportion 

cos.  M  cos.  N  :  sin.  M  sin.  N  =.  y  :  x  \ 
hence,  by  the  theory  of  proportions, 

cos.  M  cos.  N  —  sin.  M  sin.  N  :  cos.  M  cos.  N 
-\-  sin.  M  sin.  iV  =  y  —  x  \x  -\-  y, 
or,  by  (28)  and  (29), 

cos.  (ilf  +  iV)  :  cos.  (iKf—  N)  =z  y  —  x\y  +  x% 
which  may  be  written  as  in  (290). 

78.  Theorem,  The  sine  of  half  the  sum  of  two  angles 
of  a  spherical  triangle  is  to  the  sine  of  half  their  differ  - 
ence7  as  the  tangent  of  half  the  side  to  which  they  are 

16 


(292) 


93) 


182  SPHERICAL    TRIGONOMETRY.  [CH.  III. 

INeper's  Analogies. 

both  adjacent  is  to  the  tangent  of  half  the  difference  of 
the  other  two  sides ;  that  is;  in  the  spherical  triangle 
AB C  (figs.  32  and  33.), 
sin.J  (A+C) :  sin.  £  (A— C)=  tang.  J  b  :  tang.  \  {a—c).  (291) 

Proof  The  quotient  of  (284),  divided  by  (286)  is,  by  an 
easy  reduction, 

tang,  i  A sin.  (s  —  c) 

tang.  J  C        sin.  (s  —  a) 

Hence,  by  §  76, 

sin.  £  (A  +  C)  sin.  (5  —  c)  -f-  sin.  (s  —  a) 

sin.  %(A  —  C)        sin.  (s  —  c)  — sin.  (s  —  a)' 

If  we  make  in  equation  (40) 

A  =z  s  —  c  =z  J  (a  -f-  b  —  c), 
B  =  s  —  a  =  i(—  a  +  b  +  c); 
we  have 

A  +  B  =  b, 

A  —  B  —  a  —  c; 

and  (40)  becomes 

sin.  (s  —  c)  4-  sin.  (s  —  a)  _  tang.  £  b 

sin.  (s  —  c)  —  sin.  (s  —  a)  tang.£(a — c)' 

This  equation,  substituted  in  the  second  member  of  (293), 
gives 

sin,  j  (A  +  C)  _       tang.  |  6 

sin.  J  (A  —  C)  ~  tang.  J  (a  —  c)'  V       ; 

which  is  the  same  as  (291). 

79.  Theorem.  The  cosine  of  half  the  sum  of  two 
angles  of  a  spherical  triangle  is  to  the  cosine  of  half 


<§>  79.]  SPHERICAL    OBLIQUE    TRIANGLES.  183 

Neper's  Analogies. 

their  difference,  as  the  tangent  of  half  the  side  to  which 
they  are  both  adjacent  is  to  the  tangent  of  half  the  sum 
of  the  other  two  sides ;  that  is,  in  the  spherical  triangle 
iiflC(figs.  32  and  33.), 

cos.i(A+C):coa.£(A— C)—  tang.^-6  :tang.  J(«  +  c).  (295) 

Proof.  The  product  of  (284)  and  (286)  is,  by  a  simple  re- 
duction, 

„       sin.  (s  —  b) 
tang,  i  4;tang.  £  C  =  -   ^  g       J 

hence,  by  §  77, 

cos.  £  (A  -f-  C )        sin.  s  —  sin.  (s  —  b) 


cos.  £  (A  —  C)        sin.  5  -\-  sin.  (s  —  b) 
If  in  equation  (40)  inverted  we  make 
A  =  s  =  i  (a  +  b  +  c), 

we  have 

A  +  JB  r=  a  +  c, 

4  —  J3  =  6  ; 
and  (40)  becomes 

sin.  s  —  sin.  (s  —  b)  tang.  J  b 

sin.  5  -)-  sin.  (s  —  b)        tang.  £  (a-(-c)" 

This  equation,  substituted  in  (296),  gives 

cos.^(A  +  C)  __       tang,  j  b 
cos.  £  (A —  C)        tang,  ^-(a-f-c)' 

which  is  the  same  as  (295). 


(296) 


(297) 


184  SPHERICAL    TRIGONOMETRY.  [CH.   III. 

Neper's  Analogies. 

80.  Scholium.    In  using  (291)  and  (295),  the  signs  of  the 
terms  must  be  attended  to  by  means- of  PL  Trig.  §  61. 


81.    Examples. 

1.  Given  in  a  spherical  triangle  two  angles  z=  158°,  and  : 
98°,  and  the  included  side  ar  144°  ;  to  find  the  other  sides. 


Solution.    By  (291), 

}  (A  +  C)  =  128° 

sin.  (ar.co.)  10.10347 

t(A  —  C)=    30° 

sin.                   9-69897 

i    b              =12° 

tang.                0.48822 

£  (a  —  c)   ==    62°  53'  1"  tang.  0.29066 

By  (295), 

£  (A  +  C)  ==  128°  cos.  (ar.  co.)  10.21066  n. 

%(A—C)=z    30°  cos.  9.93753 

i    b  =    72°  tang.  0.48822 


i  (a  +  c)  =z  103°  0'  25"     tang.  0.63641  n. 

Ans.     a  =  165°  53'  26", 
c  =    40°    7'  24". 

2.  Given  in  a  spherical  triangle  two  angles  zs  170°,  and  : 
2°,  and  the  included  side  —  92° ;  to  find  the  other  sides, 

Ans.     a  ==  103°    6'  30", 
c  =    11°  1730". 


§  85,]  SPHERICAL    OBLIQUE    TRIANGLES.  185 

Three  angles  given. 

82.  Problem.  To  solve  a  spherical  triangle,  when  its 
three  angles  are  given. 

Solution.  If  A,  B,  C  are  the  angles  of  the  given  triangle, 
and  a,  b,  c  its  sides  180°  —  A,  180°  —  By  180°  —  C  are  the 
sides  of  the  polar  triangle,  and  180°  —  a,  180°  —  6,  180°  — c 
the  angles  of  the  polar  triangle,  the  sides  are  then  given  in  the 
polar  triangle  ;  to  find  the  angles.  For  this  purpose  we  may 
use  the  formulas  of  the  preceding  problem. 

83.  Corollary.  Applying  (272)  to  the  polar  triangle  gives 

cos.  C+  cos.  A  and  B  /nA0. 

COS.  C  — rJ—/ : : ^ .  (298) 

sin.  A  sin.  B 

84.  Corollary.  Equations  (276-278)  give,  for  the  polar  tri- 
angle, if  we  put 

8  =  i  (A  +  B  +  C),  (299) 

if  we  use  (71  and  72), 

,/ — cos.  S  cos.  (S — A)\        /rt/w,v 

sin.  i  a  -  s/  ( r      n    .-^ L  1 ,      (300 

2  \  sin.  B  sin.  C  J 

•      i  i  //—  cos- S  cos.  (S—B)\        /qm, 

sin.  i  b  =V( : 7— — ?t I?      (301) 

2  \  sin.  A  sin.  C  / 

sin.  ic  =  v| : t-^td I-       (302) 

2  \  sin.  A  sin.  I*  / 

85.  Corollary.  Equations  (281-283),  applied  to  the  polar 
triangle,  give 

,/cos.  (S— B)cos.(S—  C)\ 

cos.  i  a  =  \/( Wl — ET^T^ '  I  >        (303) 

2  \  sm.  B  sin.  C  / 

,/cos.  (#  — yl)cos.(#— • C)\         /OAy4X 

cos.  J  6  =  s/ 1 i^ /-.— ^ -  1 ,       (304) 

2  \  sin.  A  sin.  C  / 

16* 


186  SPHERICAL    TRIGONOMETRY.  [CH.  III. 

Three  angles  given. 


COS. 


ic  =  vC°5-{8-AlC0S^-B)).       (305) 
2  v  \  sm.  A  sm.  B  t 


86.  Corollary.    Equations  (284-286),  applied  to  the  polar 
triangle,  give 

//  —  cos.tfcos.  (S—  A)     \ 

,/    — cos.  S  cos.  (S — B)     \         ,OAWv 

taQg-  * h  =  fi^J^^^rcg  •     <307> 

,/    — cos.  #cos.  (S — C)     \         /OAOv 
tang,  i  c  =  V(C0S-  is_AT<tis^B)y       (308) 

87.  Corollary.    Equation  (273),    applied  to  the  polar  tri- 
angle, is 

~/.  vo        — cos.  C — cos.  (A4-B)  /r»™v 

2   sin.  £  c)*  H: r— A    .    vp^     ',  (309 

v        *    '  sin.  ^1  sin.  B 

which  may  be  used  like  equation  (279). 

88.    Example. 

Given  in  the  spherical  triangle  ABC,  the  three  angles  equal 
to  89°,  5°,  and  88°  ;  to  solve  the  triangle. 

Ans.     The  three  sides  are  53°  10',  4°,  and  53°  8'. 

89.  Theorem.  The  sine  of  half  the  sum  of  two  sides 
of  a  spherical  triangle  is  to  the  tangent  of  half  their 
difference,  as  the  cotangent  of  half  the  included  angle 
is  to  the  tangent  of  half  the  difference  of  the  other  two 
angles,  that  is,  in  ABC  (figs.  32  and  33.), 
sin.|(a -f-  c) :  sin.  £(a—c)=z  cotan.  J  B  :  tang.  £(A  —  C)  (310) 


<§>  92.]  SPHERICAL    OBLIQUE    TRIANGLES.  187 

Neper's  Analogies. 

Proof.  This  theorem  is  at  once  obtained  by  applying  §  78 
to  the  polar  triangle. 

90.  Theorem.  The  cosine  of  half  the  sum  of  two  sides 
of  a  triangle  is  to  the  cosine  of  half  their  difference,  as 
the  cotangent  of  half  the  included  angle  is  to  the  tangent 
of  half  the  sum  of  the  other  two  angles ,  or  in  (figs.  32 
and  33.), 

cos.£(a-\-c(  :  cos.4-(ez  —  c)  —  cota.n.£  B  :tzng.£(A-\-C).  (311) 

Proof.  This  theorem  is  at  once  obtained  by  applying  §  79 
to  the  polar  triangle 

91.  Corollary.  These  two  theorems,  similar  to  <§>  78 
and  79,  were  given  by  Neper  for  the  solution  of  the  case, 
in  which  two  sides  and  the  included  angle  are  given. 
By  means  of  them  the  other  two  angles  can  be  found 
without  the  necessity  of  calculating  the  third  side.  In 
using  them  regard  must  be  had  to  the  signs  of  the  terms 
by  means  of  PI.  Trig.  §  61. 

92.    Examples. 

1.  Given  in  a  spherical  triangle  two  sides  z=   149°,  and 

—  49°,   and  the  included  angle  =  88° ;  to  find  the  other  an- 
gles. 

Solution.    By  §  89, 

£(a+  c)  =  99°  sin.  (ar.  co.)  10.00538 

£  (a  —  c)  =  50°  sin.  9.88425 

£B             z=U°  cotan.  0.01516 


£  {A  —  C)  —  38°  46'  10"      tang.  9.90479 


188  SPHERICAL    TRIGONOMETRY.  [CH.  III. 

Neper's  Analogies. 

By  §  90, 

£  (a  +  c)  =    "°  cos.  ar.  co.)  10.80567  n. 

i(a—  c)  =    50°  cos.  9.80807 

£B  =    44°  cotan.  0.01516 


i  (A  +  C)  =  103°  12'  31"      tang.  0.62890  n. 

Arts.     A  ~  141°  59'  41", 

Cz:    64°  27' 21". 

2.  Given  in  a  spherical  triangle  two  sides  ca  13°,  and  =  9°, 
and  the  included  angle  a:  176°  ;  to  find  the  other  angles. 

Arts.     2°  24'  7",  and  1°  40'  13". 


SPHERICAL    ASTRONOMY. 


SPHERICAL    ASTRONOMY. 


CHAPTER   I. 

THE    CELESTIAL    SPHERE    AND    ITS    CIRCLES. 

1.  Astronomy  is  the  science  which  treats  of  the 
heavenly  bodies. 

2.  Mathematical  Astronomy  is  the  science  which 
treats  of  the  positions  and  motions  of  the  heavenly 
bodies. 

The  elements  of  position  of  a  heavenly  body  are  (Geo.  §  8) 
distance  and  direction. 

3.  Spherical  Astronomy  regards  only  one  of  the  ele- 
ments of  position,  namely,  direction,  and  usually  refers 
all  directions  to  the  centre  of  the  earth. 

4.  In  spherical  astronomy  all  the  stars  may,  then,  be 
regarded  as  at  the  same  distance  from  the  earth's  centre 
upon  the  surface  of  a  sphere,  which  is  called  the  celes- 
tial sphere. 

Upon  this  imaginary  sphere  are  supposed  to  be  drawn  vari- 
ous circles,  which  are  divided  into  the  well  known  classes  of 
great  and  small  circles.  [B.  p.  47.] 


192  SPHERICAL    ASTRONOMY.  [CH.  I 

Secondaries.  Declination.  Hour  circles. 

"  All  angular  distances  on  the  surface  of  the  sphere,  to  an 
eye  at  the  centre,  are  measured  by  arcs  of  great  circles." 
[B.  p.  48.] 

5.  "  Secondaries  to  a  great  circle  are  great  circles 
which  pass  through  its  poles,  and  are  consequently  per- 
pendicular to  it."  [B.  p.  48.] 

6.  "If  the  plane  of  the  terrestrial  equator  be  pro- 
duced to  the  celestial  sphere,  it  marks  out  a  circle  called 
the  celestial  equator ;  and  if  the  axis  of  the  earth  be 
produced  in  like  manner,  it  becomes  the  axis  of  the 
celestial  sphere ;  and  the  points  of  the  heavens,  to 
which  it  is  produced,  are  called  the  poles,  being  the 
poles  of  the  celestial  equator." 

"  The  star  nearest  to  each  pole  is  called  the  pole 
star."  [B.  p.  48.] 

7.  "  Secondaries  to  the  celestial  equator  are  called 
circles  of  declination;  of  these  24,  which  divide  the 
equator  into  equal  parts  of  15°  each,  are  called  hour 
circles" 

"  Small  circles,  parallel  to  the  celestial  equator,  are 
called  parallels  of  declination."  [B.  p.  48.] 

The  parallels  of  declination  correspond,  therefore,  to  the 
terrestrial  parallels  of  latitude,  and  the  circles  of  declination 
to  the  terrestrial  meridians.  A  certain  point  of  the  celestial 
equator  has  been  fixed  by  astronomers,  and  is  called  the  vernal 
equinox.  The  circle  of  declination,  which  passes  through  the 
vernal  equinox,  bears  the  same  relation  to  other  circles  of 


$  11.]  CELESTIAL    SPHERE    AND    ITS    CIRCLES.  193 

Right  ascension.  Horizon. 

declination,  which  the  first  meridian  does  to  other  terrestrial 
meridians. 

8.  "  The  declination  of  a  star  is  its  angular  distance 
from  the  celestial  equator,"  measured  upon  its  circle  of 
declination.  [B.  p.  49.] 

9.  The  right  ascension  of  a  star  is  the  arc  of  the 
equator  intercepted  between  its  circle  of  declination 
and  the  vernal  equinox.   [B.  p.  49.] 

Right  ascension  is  either  estimated  in  degrees,  minutes,  &c. 
from  0°  to  360° ;  or  in  hours,  minutes,  &,c.  of  time,  15  de- 
grees being  allowed  for  each  hour,  as  in  Sph.  Trig.  §  3. 

The  positions  of  the  stars  are  completely  determined  upon 
the  celestial  sphere,  when  their  right  ascensions  and  declina- 
tions are  known.  Catalogues  of  the  stars  have  accordingly 
been  given,  containing  their  right  ascensions  and  declina- 
tions. [B.  Table  viii.  p.  80.] 

10.  "  The  sensible  horizon  is  that  circle  in  the  heav- 
ens, whose  plane  touches  the  earth  at  the  spectator." 

"  The  rational  horizon  is  a  great  circle  of  the  celes- 
tial sphere  parallel  to  the  sensible  horizon."  [B.  p.  48.] 

11.  The  radius,  which  is  drawn  to  the  observer,  is 
called  the  vertical  line. 

The  point,  where  the  vertical  line  meets  the  celestial 
sphere  above  the  observer,  is  called  the  zenith  ;  the  op- 
posite point,  where  this  line  meets  the  sphere  below  the 
observer,  is  called  the  nadir. 
17 


194  SPHERICAL    ASTRONOMY.  [CH.  I. 

Prime  vertical.  Cardinal  points. 

Hence  the  vertical  line  is  a  radius  of  the  celestial  sphere 
perpendicular  to  the  horizon ;  and  the  zenith  and  nadir  are 
the  poles  of  the  horizon.    [B.  p.  48.] 

12.  Circles  whose  planes  pass  through  the  vertical 
line  are  called  vertical  circles.     [B.  p.  48.] 

The  vertical  circles  are  secondaries  to  the  horizon. 

13.  The  vertical  circle  at  any  place,  which  is  also  a 
circle  of  declination,  is  called  the  celestial  meridian  of 
that  place.    [B.  p.  48.] 

The  plane  of  the  celestial  meridian  of  a  place  is  the  same 
with  that  of  the  terrestrial  meridian. 

14.  The  points,  where  the  celestial  meridian  cuts 
the  horizon,  are  called  the  north  and  south  points. 
[B.  p.  48.] 

The  north  point  corresponds  to  the  north  pole,  and  the  south 
point  to  the  south  pole. 

15.  The  vertical  circle,  which  is  perpendicular  to 
the  meridian,  is  called  the  prime  vertical.    [B.  p.  48.] 

16.  The  points,  where  the  prime  vertical  cuts  the 
horizon,  are  called  the  east  and  west  points.   [B.  p.  48.] 

"To  an  observer,  whose  face  is  directed  towards  the  south, 
the  east  point  is  to  his  left  hand,  and  the  west  to  his  right 
hand.  Hence  the  east  and  west  points  are  90°  distant  from 
the  north  and  south.  These  four  are  called  the  cardinal 
points." 


"§>   18.]  CELESTIAL    SPHERE    AND    ITS    CIRCLES. 


195 


Altitude. 


Azimuth. 


H  The  meridian  of  any  place  divides  the  heavens  into  two 
hemispheres,  lying  to  the  east  and  west;  that  lying  to  the  east 
is  called  the  eastern  hemisphere,  and  the  other  the  western 
hemisphere." 

17.  The  altitude  of  a  star  is  its  angular  distance  from 
the  horizon,  measured  upon  the  vertical  circle  passing 
through  the  star.  [B,  p.  48.] 


18.  The  azimuth  of  a  star  is  the  arc  of  the  horizon 
intercepted  between  its  vertical  circle  and  the  north  or 
south  point.    [B.  p.  48.] 

A  star  may  be  found  without  difficulty,  when  its  altitude 
and  azimuth  are  known.  But  these  elements  of  position  are 
constantly  varying. 


196 


■■*- 


,  Al,     -VSTRONOMY. 


[CH  II. 


rixrtl  stars. 


Planets. 


Constellations. 


OllAPTKR    1L 


TIIK    niruNvi,    MOTION. 


19.  "  Stars  are  distinguished  into  two  kinds,  fixed 
and  wandering."     [H.  p.    15. J 

Most  of  the  stars  are  fixed,  thai  is,  retain  constantly  almost 
the  same  relative  position  ;  so  that  the  same  celestial  globes 
and  maps  continue  to  be  accurate  representations  of  the  fir- 
mament for  many  years.  This  is  a  fact  of  fundamental  impor- 
tance, and  furnishes  the  fixed  points  for  arriving  at  a  complete 
knowledge  of  the  celestial  motions.  Small  changes  of  position 
have,  indeed,  been  detected  even  in  the  fixed  stars,  as  will  be 
shown  in  tho  course  of  this  treatise;  but  these  changes  are 
too  small  to  disturb  the  general  fact  ;  they  are,  indeed,  too 
small  ever  to  have  been  detected,  if  the  positions  of  the  stars 
had  been  subject  to  great  variations. 

20.  Of  the  wandering  stars  there  are  eleven,  which 
are  called  planets.  They  are  Mercury  (  g  )>  Venus  (  ?  ), 
the  Earth  (0),  Mars  (  $  ),  Vesta  (g),  Juno  ($  ),  Pal- 
las ($),  Ceres  (?),   Jupiter  (#),    Saturn  (h),   and 

Cranus  ( * ).     [B.  p.  45.] 


21.    For  the   sake    of  remembering    the  stars  with 
greater  ease,  they  have  been  divided  into  groups  called 

constellations  ;  and  to  uive  distinctness  to  the  constella- 
tions, they  have  been  supposed  to  be  circumscribed  by 


$  22.]  DIURNAL    MOTION.  197 

Diurnal  motion. 

the  outlines  of  some  figure  which   they  were  imagined 
to  resemble.     [B.  p.  45.J 

The  stars  have  also  been  distinguished  according  to 
their  brilliancy,  as  of  the^rs^,  second,  &c.  magnitude. 

Proper  names  have  been  given  to  the  constellations 
and  to  the  most  remarkable  stars. 

The  catalogues  and  the  maps  of  the  stars  are  now  so  accu- 
rate, that  no  new  star  could  appear  without  being  detected ; 
and  any  change  in  the  place  of  any  of  the  larger  stars  would 
be  immediately  discovered. 

22.  All  the  stars  appear  to  have  a  common  motion, 
by  which  they  are  carried  round  the  earth  from  east  to 
west  in  24  hours.  This  rotation  of  the  heavens,  or  of 
the  celestial  sphere,  is  called  the  diurnal  motion. 

By  its  diurnal  motion,  the  celestial  sphere  rotates,  with  the 
most  perfect  uniformity,  about  its  axis.  The  pole  star  would, 
therefore,  if  it  were  exactly  at  the  pole,  remain  stationary  ; 
but  since  it  is  not  exactly  at  the  pole,  it  revolves  in  a  very 
small  parallel  of  declination  about  the  stationary  pole. 

Any  star  in  the  equator  revolves  in  the  plane  of  the  equator, 
and  all  other  stars  revolve  in  the  planes  of  the  parallels  of 
declination  in  which  they  are  situated. 

If  O  (fig.  34.)  is  the  place  of  the  observer,  NES  W  his 
horizon,  Z  his  zenith,  P  and  P'  the  poles,  the  star  which  is 
at  the  distance  from  P, 

PM  =  PM! 

will  appear  to  describe  the  circumference  MH'  M'  H.     It  will 
rise  in  the  east  at  H  and  set  at  H',  if  the  distance  PM' 
17* 


198  SPHERICAL   ASTRONOMY.  [CH.  II. 

Sideral  time. 

from  the  pole  is  greater  than  the  altitude  PN  of  the  pole. 
But  if  its  distance  from  the  pole 

PL  ==  PL' 

is  less  than  PN,  the  star  will  not  set,  but  will  describe  a 
circle  above  the  horizon  ;  and  if  its  distance  from  the  pole 

PG  =  PG< 

is  greater  than  the  greatest  distance  PS  from  the  pole  to  the 
horizon,  the  star  will  never  rise  so  as  to  be  seen  by  the  ob- 
server at  O,  but  will  describe  a  circle  below  the  horizon. 

23.  The  time  which  it  takes  a  star  to  pass  from  any 
position  round  again  to  the  same  position  is  called  a 
sideral  day,  that  is,  literally,  a  star-day.  This  day  is 
divided  into  24  hours,  and  clocks  regulated  to  this  time 
are  said  to  denote  sideral  time.    [B.  p.  147.] 

24.  Each  point  of  the  celestial  equator  passes  the 
meridian  once  in  a  sideral  day ;  and  the  arc  contained 
between  two  hour  circles  passes  it  in  a  sideral  hour. 
The  sideral  time,  therefore,  which  has  elapsed  since 
the  vernal  equinox  was  upon  the  equator,  is  equal  to 
the  right  ascension  of  the  meridian  expressed  in  time. 
[B.  p.  208.] 

The  meridian  changes  its  right  ascension  at  each  instant, 
precisely  as  if  the  celestial  sphere  were  stationary,  while  the 
observer,  with  his  meridian  and  zenith,  is  carried  uniformly 
round  the  earth's  centre  from  west  to  east  once  in  a  sideral 
day. 


<§>  28.]  DIURNAL    MOTION.  199 

Hour  angle.  Amplitude. 

25.  The  angle  MPB  (fig.  35.),  which  the  circle  of 
declination  of  the  star  makes  with  the  meridian,  is 
called  its  hour  angle. 

While  the  star  moves  from  the  point  M  in  the  meridian  to 
the  point  B  with  an  uniform  motion,  the  arc  MP  is  carried 
to  the  position  PB,  and  the  angle  MPB  is  described  with  an 
uniform  motion.  This  angle  converted  into  time  is,  then,  the 
sideral  time  since  the  passage  of  the  star  over  the  meridian. 

26.  Corollary.  The  difference  of  the  right  ascensions  of  the 
star  and  of  the  meridian  is  the  hour  angle  of  the  star. 

27.  The  distance  of  a  star  from  the  east  or  west 
points  of  the  meridian,  at  the  time  of  its  rising  or 
setting,  is  the  amplitude  of  the  star.     [B.  p.  48.] 

28.  ^Problem.  To  find  the  altitude  and  azimuth  of  a 
star ,  when  its  declination  and  hour  angle  are  known, 
and  also  the  latitude  of  the  place. 

Solution.  If  P  (fig.  35.)  is  the  pole,  Z  the  zenith,  and  3 
the  star ;  we  have 

PZ  —  polar  dist.  of  zenith  ==  co.  latitude  =  90°  —  L, 
PN  =  90°  —  PZ—L, 

PB  =  polar  dist.  of  star  =  p, 

=  co.  declination  of  star,  when  it  is  on  the  same  side 
of  the  equator  with  the  pole. 

=  90°  -)-  declination  of  star,  when  it  is  on  the  differ- 
ent side  of  the  equator  from  the  pole. 

=  90°  q=  D, 


200  SPHERICAL    ASTRONOMY.  [CH.   II. 

To  find  a  star's  altitude  and  azimuth. 

ZB  =  zenith  dist.  of  star  =  z, 

==  co.  altitude  of  star,  when  it  is  above  the  horizon. 

=  90°  -f-  depression  of  star,  when   it  is   below  the 
horizon. 

ZPB  =  #'s  hour  angle  =  h, 

PZB  =  azimuth  of  star  counted  from  the  direction  of  the 
elevated  pole. 

=  a  =  azimuth,  when  less  than  90°, 

=  180°  —  azimuth,  when  greater  than  90°. 

There  are,  then,  given  in  the  spherical  triangle  PZB,  the  two 
sides  PZ  and  PB,  and  the  included  angle  ZPB ;  so  that 
the  side  BZ  and  the  angle  PZB  can  be  calculated  by  Sph. 
Trig.  §  44. 

If  we  let  fall  the  perpendicular  BC  upon  PZ> 

tang. PC—  cos. h  tang.  (90°  =pD)==p  cos.  A  cotan.  D  (312) 

CZ=  PZ—  PC—  90°  —  (L  +  PC) 

or  —PC—  PZ=(L  +  PC)—  90°.  (313) 

Hence,  by  (241), 

cos.  PC :  sin.  (L  +  PC)  z=  d=  sin.  Z>  :  cos.  z ;        (314) 

in  which  formulas  the  upper  sign  is  used  when  the  star  is  upon 
the  same  side  of  the  equator  with  the  elevated  pole,  that  is, 
when  D  and  L  are  of  the  same  name ;  and,  by  (242), 

sin.  PC :  cos.  (L  +  PC)  —  cotan.  h  :  cotan.  a.       (315) 

29.  Corollary,    When  the  altitude  and  azimuth  are  both  to 
be  found,  the  calculation  by  the  above  method  is  as  short  as 


<§>  30.]  DIURNAL    MOTION.  201 


To  find  a  star's  altitude. 


by  any  other;  but  when,  as  is  usually  the  case,  the  altitude 
only  is  required,  the  following  method  is  preferable. 

We  have 

PZ  +  PB  =  180°  —  L  =f  D  =  180°  —  (L  ±  D) 

PB  —  PZ=^D  +  L  =  L^D)*, 

whence,  by  (249)  and  (250), 

cos.z  =  —  cos.  (L^zD)-{-2cos.Dcos.L(cos.^h)2    (316) 
cos.z  =      cos.(Z=pZ>) — 2cos.  ZJcos.  Z,(sin.  J/*)2,    (317) 

which  may  be  used  at  once,  and  {317)  may  be  calculated  by 
the  aid  of  the  column  of  Rising  in  Table  XXIII.  The  rule 
obtained  from  (317)  is  the  same  with  that  on  p.  250  of  the 
Navigator,  remembering  that  when  the  star  is  above  the  hori- 
zon 

cos.  z  s=  sin.  ^c's  alt.  (318) 

But  when  the  star  is  below  the  horizon 

cos.  z  z=  —  sin.  sjc's  depression.  (310) 

30.  Corollary.  If  the  given  hour  angle  is  6h  =  90°,  the 
problem  is  at  once  reduced  to  the  solution  of  a  right  triangle. 
We  in  this  case  have,  by  Napier's  Rules, 

cos,  z  ==  sin.  L  cos.  p, 

or  sin.  ^c's  alt.  z=  ±  sin.  L  sin.  D  (320) 

cotan.  a  z=  cos.  L  cotan.  p 

cotan.  sfc's  azimuth  ==  dc=  cos.  L  tang.  D.         (321) 

The  upper  sign  is  to  be  used  in  formulas  (320)  and  (321), 
when  the  declination  is  of  the  same  name  with  the  latitude; 
otherwise  the  lower  sign.     In  the  former  case,  therefore,  the 


202  SPHRICAL    ASTRONOMY.  [CH.  II. 


To  find  a  star's  altitude  and  azimuth. 


star  is  above  the  horizon  when  its  hour  angle  is  six  hours,  and 
on  the  same  side  of  the  prime  vertical  with  the  elevated  pole  ; 
but,  in  the  latter  case,  it  is  below  the  horizon,  and  on  the  same 
side  of  the  prime  vertical  with  the  depressed  pole. 

31.  Corollary,  If  the  star  is  in  the  celestial  equator,  as  in 
(fig.  36.),  we  have  in  the  right  triangle  BZQ} 

BQ  ==  BPQ  —  h 

ZQ  —  L 

QZB  =  180°  —  a, 

whence  cos.  z  t=  cos.  L  cos.  h, 

or  sin.  ^c's  alt.  =  cos.  L  cos.  h  (322) 

cotan.  (180°  —  a)  s±  sin.  L  cotan.  ft, 

or  cotan.  a  =  —  sin.  L  cotan.  h,  (323) 

Hence,  if  the  hour  angle  is  less  than  six  hours,  the  star  which 
moves  in  the  celestial  equator  is  above  the  horizon,  and  on  the 
same  side  of  the  prime  vertical  with  the  depressed  pole  ;  but 
if  the  hour  angle  is  greater  than  six  hours,  this  star  is  be- 
low the  horizon,  and  on  the  same  side  of  the  prime  vertical 
with  the  elevated  pole. 

32.  Corollary.  If  the  place  is  at  the  equator,  as  in  (fig.  37.), 
the  celestial  equator  ZE  is  the  prime  vertical,  so  that  if  the 
hour  circle  PB  is  produced  to  C,  we  have  in  the  right  tri- 
angle ZBC 

ZC—  ZPBz=  h 
BZC  =  90°  —  a 
BC—  D, 


§  33.]  DIURNAL    MOTION.  203 


To  find  a  star's  altitude  and  azimuth. 

whence  cos.  z  =  cos.  D  cos.  h, 

or  sin.  2(c's  alt.  —  cos.  D  cos.  h  (324) 

cotan.  (90°  —  a)  txz  sin.  h  cotan.  D, 
or  tang,  a  ==  sin.  h  cotan.  D  ;  (325) 

so  that  the  star  is  above  the  horizon  when  the  hour  angle  is 
less  than  six  hours,  and  below  the  horizon  when  the  hour 
angle  is  greater  than  six  hours. 

33.  Examples. 

1.  Find  the  altitude  and  azimuth  of  Aldebaran  to  an  ob- 
server at  Boston,  in  the  year  1830,  when  the  hour  angle  of  this 
star  is  3*  25m  12s. 

Solution.    We  find  by  tables  VIII  and  LIV 

D—  16°  11' N.  L  —  42°21'N. 

Hence 

h  =  3*  25 m  12*  log.  col.  Ris.  4.57375 

L  ts  42°  21'                        cos.  9.86867 

D=i  16°  11;                        cos.  9.98244 


26599  4.42486 

■  D  =  26°  10;     nat.  cos.         89752 


alt.  —  39°  10'     nat.  sin.         63153     sec.  10.11052 

h  —  51°  18'  sin.     9.89233 

D.     cos.    9.98244 


azimuth  from  South  =  75°  10'  sin.    9.98529 


204  SPHERICAL    ASTRONOMY.  [CH.  II, 

To  find  a  star's  altitude  and  azimuth. 

2.  Find  the  altitude  and  azimuth  of  Aldebaran  at  Boston* 
in  the  year  1830,  six  hours  after  it  has  passed  the  meridian. 

Solution.  By  formulas  (320)  and  (321), 

L  =  42°  21'         sin.     9.82844         cos.     9.86867 
D  =  16°  IV         sin.     9.44516        tang.  9.46271 


alt.  =  10°  49'        sin.     9.27360 


azimuth  from  North  =  77°  54'  cotan.  9.33138 

3.  Find  the  altitude  and  azimuth  of  a  star  in  the  celestial 
equator,  to  an  observer  at  Boston,  when  the  hour  angle  of  the 
star  is  3*25™  12  s. 

Solution.    By  formulas  (322)  and  (323), 

L  =  42°  21'         cos.     9.86867         sin.     9.82844 
h  z=  51°  18'         cos.     9.79605       cotan.  9.90371 


alt.  =  27°  317         sin.     9.66472 


azimuth  from  South  =  61°  39'  cotan.  9.73215 

4.  Find  the  altitude  and  azimuth  of  Aldebaran  to  an  ob- 
server at  the  equator,  in  the  year  1830,  when  the  hour  angle 
of  the  star  is  3*25™  12s. 

Solution.    By  formulas  (324)  and  (325), 

Z>z=16°ll'         cos.     9.98244      cotan.  10.53729 
h  =  51°  18'         cos.     9.79605       cotan.     9.90371 


alt.  =  36°  54'         sin.     9.77849 


azimuth  from  North  =  70°  5'  tang.  10.44100 


§  33.]  DIURNAL    MOTION.  205 

Altitude  and  azimuth. 

5.  Find  the  altitude  and  azimuth  of  Fomalhaut  to  an  ob- 
server at  Boston,  in  the  year  1840,  when  its  hour  angle  is 
2*3^20*. 

Ans.     Its  altitude  .  .         =11°  51'. 

Its  azimuth  from  the  South     ==  15°  24'. 


6.  Find  the  altitude  and  azimuth  of  Dubhe  to  an  observer 
at  Boston,  in  the  year  1840,  when  its  hour  angle  is  9h  30  m. 

Ans.     Its  altitude  .  .         —  19°  14'. 

Its  azimuth  from  the  North     =z  17°  15'. 

7.  Find  the  altitude  and  azimuth  of  Fomalhaut  to  an  ob- 
server at  Boston,  in  the  year  1840,  when  its  hour  angle  is  6\ 

Ans.     Its  depression  below  the  horizon  t=z  19°  58'. 

Its  azimuth  from  the  South  =  38°  31'. 

8.  Find  the  altitude  and  azimuth  of  Dubhe  to  an  observer 
at  Boston,  in  the  year  1840,  when  its  hour  angle  is  6\ 

Ans.     Its  altitude  .  .  =  36°  44'. 

Its  azimuth  from  the  North      =  69°    3'. 

9.  Find  the  altitude  and  azimuth  of  a  star  in  the  celestial 
equator  to  an  observer  at  Stockholm,  when  its  hour  angle  is 
2*3"'  20*. 

Ans.     Its  altitude  .  .  ==  25°  58'. 

Its  azimuth  from  the  South      =  34°  45'. 

10.  Find  the  altitude  and  azimuth  of  a  star  in  the  celestial 

18 


206  SPHERICAL    ASTRONOMY.  [CH.  II. 

Altitude  of  a  star  in  the  prime  vertical. 

equator,  to  an  observer  at  Stockholm,  when  the  hour  angle  is 
9*  30m. 

Ans.     Its  depression  below  the  horizon  ■=  23°  51'. 
Its  azimuth  from  the  North  =  41°  44'. 


11.  Find  the  altitude  and  azimuth  of  Fomalhaut,  to  an  ob- 
server at  the  equator,  in  the  year  1840,  when  its  hour  angle 
is2*3w20*. 

Ans.     Its  altitude  .  .  —  47°  45'. 

Its  azimuth  from  the  South      =  41°    4'. 

12.  Find  the  altitude  and  azimuth  [of  Dubhe,  to  an  observ- 
er at  the  equator,  in  the  year  1840,  when  its  hour  angle  is 
9*30m. 

Ans.     Its  depression  below  the  horizon  =  21°  24'. 
Its  azimuth  from  the  North  —  17°  30'. 

34.  In  the  triangle  ZPB  (fig.  2.)  other  parts  might 
be  given  instead  of  the  two  sides  ZP,  PB,  and  the 
included  angle  P,  and  the  triangle  might  be  resolved. 
Of  the  problems  thus  derived,  we  shall  only,  for  the 
present,  consider  two  cases. 

35.  Problem.  To  find  a  given  star's  hour  angle  and 
altitude,  when  it  is  upon  the  prime  vertical. 

Solution.  The  angle  PZB  is,  in  this  case,  a  right  angle, 
and  if  we  use  the  preceding  notation,  we  have 

cos.  7i  zc  cotan.  L  cotan.  p  —  ±  cotan.  L  tang.  D     (326) 

cos.  z  z=.  cos.  p  cosec.  L, 
or     sin.  &'s  alt.  =  db  sin.  D  cosec.  L  ;  (327) 


m 

§  38.]  DIURNAL    MOTION.  207 

Altitude  of  a  star  in  the  prime  vertical. 

so  that  when  the  declination  and  latitude  are  of  the  same  name, 
the  hour  angle  is  less  than  6  hours,  and  the  star  is  above  the 
horizon  ;  but  when  the  declination  and  latitude  are  of  differ- 
ent names,  the  hour  angle  is  greater  than  6  hours,  and  the 
star  is  below  the  horizon. 

36.  Scholium.  The  problem  is,  by  Sph.  Trig.  §  27,  impos- 
sible, when  the  declination  is  greater  than  the  latitude;  so 
that,  in  this  case,  the  star  is  never  exactly  east  or  west  of  the 
observer. 

37.  Scholium.  The  problem  is,  by  Sph.  Trig.  §  28,  indeter- 
minate, when  the  latitude  and  declination  are  both  equal  to 
zero ;  so  that,  in  this  case,  the  star  is  always  upon  the  prime 
vertical. 


38.    Examples. 

1.  Find  the  hour  angle  and  altitude  of  Aldebaran,  when  it 
is  exactly  east  or  west  of  an  observer  at  Boston,  in  the  year 
1840. 

Ans.     The  hour  angle  =  4A  45  w  44*. 

The  altitude       m  24°  26'. 

2.  Find  the  hour  angle  and  altitude  of  Fomalhaut,  when  it 
is  exactly  east  or  west  of  an  observer  at  Boston,  in  the  year 
1840. 

Ans.     The  hour  angle         .         .         s=  8h  40 m  50'. 

The  depression  below  the  horizon  =  48°  49'. 

3.  Find  the  hour  angle  and  altitude  of  Dubhe,  when  it  is 


208  SPHERICAL    ASTRONOMY.  [CH.  II. 

Time  of  a  star's  rising. 

exactly  east  or  west  of  an  observer  at  Boston,  in  the  year 
1840. 

Arts.     Dubhe  is  never  upon  the  prime  vertical  of  Boston. 

4.  Find  the  hour  angle  and  altitude  of  Canopus,  when  it  is 
exactly  east  or  west  of  an  observer  at  Boston,  in  the  year 
1840. 

Ans.     Canopus  is  never  upon  the  prime  vertical  of  Boston. 

39.  Problem.  To  find  the  hour  angle  and  amplitude 
of  a  star,  when  it  is  in  the  horizon. 

Solution.  In  this  case  the  side  ZB  (fig.  35.)  of  the  triangle 
ZPB  is  90°.  The  corresponding  angle  of  the  polar  triangle 
is,  therefore,  a  right  angle,  and  the  polar  triangle  is  a  right 
triangle,  of  which  the  other  two  angles  are 

180°  —  PZ  —  180°  —  (90°  —  L)  —  90°  +  Ly 
and       180°  —  PB  =  180°  —  (90°  =p  D)  —  90°  d=  D. 

The  hypothenuse  of  the  polar  triangle  is  180°  —  h,  and  the 
leg,  opposite  the  angle,  90°  ±  D,  is  180°  —  a. 

Hence,  by  Sph.  Trig.  §  40,  and  PI.  Trig.  §  60  and  62, 

—  cos.  7i  =  zh  tang.  L  tang.  Z>, 

or  cos.  h  =.  =F  tang.  L  tang.  D  (328) 

—  cos.  a  =  ^p  sin.  D  sec.  L, 

or  cos.  a  =  d=  sin.  D  sec.  L ;  (329) 

in  which  the  upper  sign  is  used  when  the  latitude  and  declina- 
tion have  the  same  name,  and  the  lower  sign  when  they  have 
different  names ;  so  that  in  the  former  case  the  hour  angle  is 
greater  than  6  hours,  and  the  azimuth  is  counted  from  the 


<§>  41.]  DIURNAL    MOTION. 


209 


Time  of  a  star's  rising. 


direction  of  the  elevated  pole  ;  but  in  the  latter  case,  the  hour 
angle  is  less  than  6  hours,  and  the  azimuth  is  counted  from 
the  direction  of  the  depressed  pole.  The  amplitude  is  the 
difference  between  the  azimuth  a  and  90°.     Hence 

cos.  ^c's  azim.  =  sin.  ^c's  amp.  =  sin.  D  sec.  L.     (330) 

40.  Scholium.  The  problem  is,  by  Sph.  Trig.  §  41,  impos- 
sible, when  the  sum  of  the  declination  and  latitude  is  greater 
than  90°  ;  so  that,  in  this  case,  the  star  does  not  rise  or  set. 


41.    Examples. 

1.  Find  the  hour  angle  and  amplitude  of  Aldebaran,  when 
it  rises  or  sets,  to  an  observer  at  Boston,  in  the  year  1840. 

Ans.     The  hour  angle  ==  7h  lm  21*. 
The  amplitude    =  22°  9'  N. 

2.  Find  the  hour  angle  and  amplitude  of  Fomalhaut,  when 
it  rises  or  sets,  to  an  observer  at  Boston,  in  the  year  1840. 

Ans.     The  hour  angle  =  Sh  5lm  18*. 
The  amplitude   —  43°  19'  S. 

3.  Find  the  hour  angle  and  amplitude  of  Dubhe,  when  it 
rises  or  sets,  to  an  observer  at  Boston,  in  the  year  1840. 

Ans.     Dubhe  neither  rises  nor  sets  at  Boston. 

4.  Find  the  hour  angle  and  amplitude  of  Canopus,  when  it 
rises  or  sets,  to  an  observer  at  Boston,  in  the  year  1840. 

Ans.     Canopus  neither  rises  nor  sets  at  Boston. 


18* 


210  SPHERICAL    ASTRONOMY.  [CH.   III. 

Determination  of  the  meridian  line. 


CHAPTER   III. 


THE    MERIDIAN. 


42.  The  intersection  of  the  plane  of  the  meridian 
with  that  of  the  horizon  is  called  the  meridian  line. 

43.  Problem.     To  determine  the  meridian  line. 

Solution.  First  Method.  Stars  obviously  rise  to  their  great- 
est altitude  in  the  plane  of  the  meridian  ;  so  that  if  their 
progress  could  be  traced  with  perfect  accuracy,  and  the  instant 
of  their  rising  to  their  greatest  height  be  observed,  the  direc- 
tion of  the  meridian  line  could  be  exactly  determined.  But 
stars,  when  they  are  at  their  greatest  height,  change  their 
altitude  so  slowly,  that  this  method  is  of  but  little  practical 
value. 

Second  Method.  A  star  is  evidently  at  equal  altitudes  when 
it  is  at  equal  distances  from  the  meridian  on  opposite  sides  of 
it.  If,  therefore,  the  direction  and  altitude  of  a  star  are  ob- 
served before  it  comes  to  the  meridian ;  and  if  its  direction  is 
also  observed,  when  it  has  descended  again  to  the  same  alti- 
tude, after  passing  the  meridian ;  the  horizontal  line,  which 
bisects  the  angle  of  the  two  horizontal  lines  drawn  in  the 
directions  thus  determined,  is  the  meridian  line. 

Third  Method.  [B.p.  147.]  The  time  which  elapses  between 
the  superior  and  inferior  passage  of  a  star  over  the  meridian  is 
just  half  of  a  sideral  day.  If,  then,  a  telescope  were  placed  so  as 


<§>  43.]  THE    MERIDIAN.  211 

Meridian  determined  by  circumpolar  stars. 

to  revolve  on  a  horizontal  axis  in  the  plane  of  the  meridian,  the 
two  intervals  of  time  between  three  successive  passages  of  a 
star  over  the  central  wire,  must  be  exactly  equal.  But  if  the 
vertical  plane  of  the  telescope  is  not  that  of  the  meridian, 
these  two  intervals  will  not  be  equal,  and  the  position  of  the 
telescope  must  be  changed  until  they  become  equal. 

Thus,  if  ZMmN  (fig.  37.)  is  the  plane  of  the  meridian, 
Z S  s  T  that  of  the  vertical  circle  described  by  the  telescope, 
MS  WsmE  the  circle  of  declination  described  by  the  star 
about  the  pole  P;  this  star  will  be  observed  at  the  points  S 
and  s  instead  of  at  the  points  M  and  m.  Now  the  star  de- 
scribes the  circle  of  declination  with  an  uniform  motion,  and 
therefore  the  arc  SP  moves  uniformly  with  the  star  around 
the  pole,  so  that  the  angle  SP  M  is  proportional  to  the  time 
of  its  description;  that  is,  the  angle  SP31,  reduced  to  time, 
denotes  the  sideral  time  of  its  description. 

Let  then 

T  =  the  sideral  time  of  describing  the  arc  SM} 
t  =  the  sideral  time  of  describing  the  arc  5  m, 
I  =  interval  from  the  observation  at  £  to  that  at  s, 
i  ==  interval  from  the  observation  at  s  to  that  at  S} 
<5  i  =  the  difference  of  these  two  intervals  ; 
we  have  then,  in  sideral  time, 

I  =  12h  —  T  —  t  =  12h  —  ( T  +  t) 
i  =  12*  +  T  +  t  =  12*  +  ( T  +  t ) 

di=ri  —  I=:2(T+t);  (331) 

so  that  if  T  and  t  were  equal  to  each  other,  and  they  are 
nearly  so  in  the  case  of  the  pole-star,  we  should  have 


2L2  SPHERICAL    ASTRONOMY.  [CH.  III. 

Meridian  determined  by  circumpolar  stars. 
di  =  4  T=  4t 

that  is,  the  time  of  describing  the  arc  MS  or  m  s  is  nearly  one 
quarter  part  of  the  difference  between  the  intervals. 

But  the  error  of  this  result  can  be  calculated  without  much 
difficulty.     For  this  purpose,  let 

L  ==  the  latitude  of  the  place,  =  90°  —  PZ, 

p  s=s  the  polar  distance  of  the  star  ==  PS  —  P s, 

a  =  the  azimuth  of  ZS  T  =  TN  =  TZN. 

The  arcs  MS  and  m  s  are  so  small,  that  they  do  not  differ 
sensibly  from  the  arcs  of  great  circles  drawn  from  S  and  s 
perpendicular  to  ZPN. 

If,  then,  in  the  two  right  triangles  PSM  and  ZSMf  PM 
and  ZM  are  the  middle  parts,  SM,  co.  SZM,  and  co.  SPM 

are  the  adjacent  parts,  so  that 

sin.  PM :  sin.  ZM  =  cotan.  SPM :  cotan.  SZM 

1  1 

tang.  SPM  '  tang.  SZM 

=  tang.  SZM :  tang.  SPM. 

But  Z3I=  ZP  —  PM  =  90°  —  L  —p 

and  the  angles  SZM  and  SPM  are  so  small,  that  they  are 
sensibly  proportional  to  their  tangents,  whence 

sin.  p  :  cos.  (p  +  L)  =  a  :  SPM,  (332) 

or         a  :  SPM  =?  sin.  p  :  cos.  p  cos.  L  —  sin.  p  sin.  L 

fats  1  :  cotan.  p  cos.  Z<  —  sin.  L9 


§  43.]  THE    MERIDIAN.  213 


Meridian  determined  by  circumpolar  stars.         Table  A  [B.  p.  151.] 

and  if  T  is  expressed  in  sideral  hours 

T.  15°  ±=  SPM  —  a  cotan.  p  cos.  L  —  a  sin.  L. 
In  like  manner,  we  find 

t .  15°  =  5  Pm  =  a  cotan.  p  cos.  L  -f-  a  sin.  L. 
Hence,  by  (331) 

(T+  t)  15°  =  JJi.15°  —  2«  cotan.pcos.  i 
a  cotan.  p  cos.  L  =  £$i.  15° 
T.  15°  =  ^  (52.15°  —  a  sin.  Z, 
£.15°  ±=  |  &\t#  -f  a  sin.  L 
a  =  J  (5  i .  15°  tang.  j9  sec.  Z.  (333) 

T  =  £  <J  £  —  £  <5 «  tang,  p  tang.  X 
£  =  £  «J  i  -}-  £  i?  £  tang,  p  tang.  L, 
so  that  the  correction  is 

I  3  i  tang,  p  tang.  L,  (334) 

which  is  to  he  added  to  the  quarter  interval  at  the  lower  tran- 
sit ;  and  to  be  subtracted  from  the  quarter  interval  at  the  upper 
transit. 

This  correction  is  proportional  to  the  quarter  interval,  so 
that  if  it  is  computed  for  any  supposed  value  of  this  interval, 
it  may  be  computed  for  any  other  interval  by  a  simple  propor- 
tion. Now  table  A,  page  151,  of  the  Navigator,  is  the  value 
of  this  correction,  when  the  interval  is  1000s.  It  may  be 
observed,  that  it  is  not  necessary  that  this  time  should  be 
sideral  time,  because  all  the  terms  of  the  values  of  T  and  t 
are  expressed  in  the  same  time,  which  may  be  that  of  the 
clock. 


214  SPHERICAL    ASTRONOM     .  [CH.  111. 

Table  B  [B.  p.  151.] 

The  azimuth  a  is  given  in  table  B,  [B.  p.  151],  and  may  be 
computed  from  the  formula  (333).  But  the  interval  in  the 
formula  is  supposed  to  be  sideral  time,  whereas  the  time  of 
the  table  is  that  called  solar  time,  to  which  clocks  are  usually 
regulated,  and  which  is  soon  to  be  described ;  all  that  need  be 
known  for  the  present  is,  that  an  interval  of  sideral  time  is 
reduced  to  solar  time  by  table  LII  of  the  Navigator,  or  by  the 
formula 

an  interval  of^solar  time  =  q^^  (335) 

an  interval  of  sideral  time 

Fourth  Method.  [B.  p.  149.]  This  method  of  determining 
the  meridian  is  by  means  of  two  known  circumpolar  stars, 
which  differ  nearly  12  hours  in  right  ascension.  The  upper 
passage  of  one  of  these  stars  is  to  be  observed,  and  the  lower 
passage  of  the  other.  Then  any  deviation  in  the  plane  of  the 
instrument  from  the  meridian,  will  evidently  produce  contrary 
effects  upon  the  observed  times  of  transit,  exactly  as  in  the 
upper  and  lower  transits  of  the  same  star.  The  time,  which 
elapses  between  the  two  observations,  will  differ  from  the  time 
which  should  elapse  by  the  sum  of  the  effects  of  the  deviation 
upon  the  two  stars.  In  the  use  of  this  method,  therefore,  the 
time  of  the  clock  must  be  known,  so  that  it  can  readily  be  re- 
duced to  sideral  time. 

The  deviations  in  the  time  of  passage  of  a  star,  correspond- 
ing to  any  azimuth,  can  be  calculated  by  means  of  equation 
(332).  For  this  formula  give  for  the  time  of  describing  the 
arc  SM 

T.  15°  e=  a  cos.  (p  -f-  L)  cosec.  p, 

or  T  —  T\  a  cos.  (p  -f-  L)  cosec.  p  ;  (336) 

which  may  be  used  if  T  is  expressed  in  sideral  seconds,  and 


§  44.]  THE    MERIDIAN.  215 

Table  C  [B.  p.  152.]  Table  XXII. 

the  arc  a  in  seconds  of  space.     But  if  T  is  expressed  in  solar 
time  we  have,  by  (335), 

T  —  0.0664846  a  cos.  (p  -f  L)  cosec.  p.  (337) 

In  the  same  way  the  value  of  t  for  an  inferior  passage  is 
found  to  be 

t  =3  0.0664846  a  cos.  (p  —  L)  cosec.  p.  (338) 

Now,  since  these  values  of  T  and  t  are  proportional  to  the 
azimuth  a,  their  values  may  be  computed  for  a  given  value  of 
the  azimuth,  as  1000",  and  arranged  in  a  table  like  Table  C, 
p.  152,  of  the  Navigator,  and  their  values  for  any  other  azi- 
muth can  be  obtained  by  a  simple  proportion. 

Fifth  Method.  [B.  p.  149.]  This  method  consists  in  observ- 
ing the  transits  of  two  stars,  which  differ  but  little  in  right 
ascension.  The  error  in  the  position  of  the  telescope  is,  in 
this  case,  equal  to  the  difference  in  the  errors  of  the  observed 
transits,  instead  of  the  sum,  as  in  the  preceding  method. 

44.  In  making  calculations  where  angles  are  intro- 
duced as  factors,  some  labor,  in  reducing  them  to  the 
same  denomination,  is  often  saved  by  means  of  a  table 
of  Proportional  Logarithms,  such  as  Table  XXII  of 
the  Navigator. 

This  table  was  particularly  designed  for  reducing  lunar  dis- 
tances, given  in  the  Nautical  Almanac,  for  every  3  hours 
to  any  intermediate  time.  It  contains,  on  this  account,  the 
logarithm  of  the  ratio  of  3  hours  to  each  angle  expressed  in 
time ;  that  is,  if  A  is  the  angle 


216  SPHERICAL   ASTRONOMY.  [CH.  III. 

Proportional  Logarithms. 

3^ 

Prop.  log.  A  =  log.  ~  =  log.3ft  —  log.^Lz=log.l80m — log.  A 

A 

—  log.  10800s  —  log.  A,  (339) 

so  that  if  A  in  the  second  member  is  reduced  to  seconds, 

Prop.  log.  A  —  4.03342  —  log.  A  in  seconds ;         (340) 

neglecting  the  right  hand  figure,  so  as  to  retain  only  four 
decimal  places.  This  agrees  with  the  explanation  of  the  table 
in  the  Introduction  to  the  Navigator  ;  and  it  is  evident  that  it 
is  immaterial  whether  the  angles,  whose  ratios  are  sought,  are 
given  in  time  or  in  degrees,  &c. 

Suppose,  now,  that  the  logarithm  of  the  ratio  of  two  angles 
is  sought,  A  and  a ;  we  have,  evidently, 

log.  —  ==  log.  A  —  log.  a  3=  Prop.  log.  a  —  Pr.  log.  A ;  (341 ) 

so  that  if  this  ratio,  which  we  will  denote  by  M„  were  known, 
and  if  a  were  known,  A  might  be  calculated  by  the  formula 

Prop.  log.  A  —  Prop.  log.  a  —  log.  M 

=  Prop.  log.  a  +  (ar.  co.)  log.  M;    (342) 

which  is,  therefore,  the  formula  for  calculating  the  value  of  A, 
given  by  the  equation 

A  —  a  M.  (343) 

Finally,  the  use  of  formula  (342)  is  facilitated  by  remem- 
bering that  the  arithmetical  complements  of  the  logarithms  of 
the  sine,  cosine,  tangent,  cotangent,  secant,  and  cosecant  of  an 
angle  are  respectively  the  logarithms  of  its  cosecant,  secant, 
cotangent,  tangent,  cosine^  and  sine. 


4  45.]  THE    MERIDIAN.  „    217 

Tables  A,  B,  C,  [B.  pp.  151  and  152.] 

45.    Examples. 

1.  Calculate  the  proportional  logarithm  of  0°  5'  45". 
Solution;    By  (340),  4.03342 

0°  5'  45"  =  345".  2.53782 


Prop.  log.  5'  45"  =z  1.4956 

2.  Calculate  the  corrections  of  tables  A  and  B,  [B.  p.  151.], 
as  in  table  XXII,  when  the  latitude  is  42°,  and  the  polar  dis- 
tance of  the  star  30°. 

Solution.  By  means  of  proportional  logarithms,  and  equations 
(333)  and  (334), 

£.1000*  =z4m  10*       Prop.  log.    1.6355  1.6355 

L  —  42°  cotan.  10.0456   cos.    9.8711 

30°  cotan.  10.2386  10.2386 


corr.  A  i=  130s  =  2m  10*     Pr.  log.   1.9197 


0.0664846  8.8227 


corr.  B=:48/41"  Prop.  log.   0.5679 

3.  Calculate  the  corrections  of  table  C  [B.  p.  152.]  for  the 
pole  star  and  the  latitude  of  30a,  when  the  polar  distance  of 
this  star  is  1°  32'  37". 

19 


218 


SPHERICAL    ASTRONOMY. 


[CH.  III. 


Determination  of  the  meridian  line. 


Solution.    By  (337)  and  (338), 

0.0G64846  8:82273 

an  1000"  3.00000 

p=l°  32'  37"  cosec.  11.56964 

p+L  =  31°  32' 37"   cos.     9.93058 
p  —  L-—  28°  26'  22" 


corr.  C  upper  trans.  =  2103 *     3.32295 
corr.  C  lower  trans.  =  2170  s 


8.82273 

3.00000 

11.56964 

9.94414 
3.33651 


4.  An  observer  at  Boston  in  the  year  1840,  wishing  to  de- 
termine his  meridian  line,  observed  three  successive^transits  of 
p  Cephei  over  the  central  vertical  line  of  his  transit  instru- 
ment, by  means  of  a*  clock  regulated  to  solar  time,  and  found 
them  to  occur  as  follows  ;  the  first  upper  transit  at  7 h  45 m  28s 
P.  M.,  the  next  inferior  transit  the  next  day  at  7h  41 m  A.M. 
the  third  transit  at  7*  41*  32  s  P.  M.  What  were  the  times  o  f 
the  star's  passing  the  meridian  the  second  day?  and  wThat  was 
the  azimuth  error  in  the  position  of  the  instrument  ? 

Solution. 

The  first  interval      s=  19*  4lw  —  7*45™  28*  =  ll*55m  32*. 

The  second  interval  =  19*  41OT  32s  —  7*  41w  =  12*   0m  32*. 


Hence 


*i  —  5m  —  300*. 


Now  L  ==  42°  21',     D  =  69°  52',    p  =z  20°  8'. 

Hence,  by  tables  A  and  B, 

corr.  A  =  83*  X  0.3  =  25', 
corr.  B  =i  31'  6"  X  0.3  =  9'  19"; 


<§>  45.]  THE    MERIDIAN.  219 

Determination  of  the  meridian  line. 

so  that  the  error  in  the  time  of  the  upper  transit  is 

£.300*  —  25*  ~  75  s  —  25*  M  50% 

and  the  error  in  the  time  of  the  lower  transit  is 

£.300*  +  25s  ±-  75s  +  25s  =  100s  —  lm  40*. 

The  times  of  the  star's  passing  the   meridian  the  second  day 
were,  then, 

7h  41™  +  lm  40s  =  T  42w  40*  A.  M. 

and  T  41-  32s  —  50s  =  lh  40w  42s  P.  M. 

The  error  in  the  azimuth  of  the  instrument  was  9'  19"  to  the 
west  of  north. 

5.  An  observer  at  Boston,  wishing  to  determine  his  meridian 
line,  on  the  morning  of  January  1,  1840,  observed,  by  means 
of  a  clock  regulated  to  solar  time,  the  superior  transit  of 
Y  UrssB  Majoris  at  5h  6m  54s  A.  M,,  and  the  inferior  transit 
of  Polaris  at  6'1  12w  23s  A.  M.  What  was  the  azimuth  error 
in  the  position  of  the  transit  instrument? 

Solution.  The  interval  between  these  two  transits  is 

6h  12™  23s  —  5h  6m  54s  =  I*  5m  29*. 

But,  by  the  Nautical  Almanac, 

12*  +  R.  A.  of  Polaris  =  13*    \m  59s 

R.  A.  of  y  Ursae  Majoris  =  1  ih  45m  25s 


Sideral  Interval         =    1*  16™  34s 


Solar  Interval  =     1*  16™  22s 

Observed  Interval     =    1*    5m  29s 


Error  of  Interval      =         10m53*  =  653*. 


220  SPHERICAL    ASTRONOMY.  [CH.  III. 

Determination  of  the  meridian  line. 

Now  for  1000"  of  azimuth   error,  and  the  latitude  of  Boston, 
Table  C  gives,    since 

Dec.  of  y  Ursae  Majoris  .         .         =  54°  35' 

Error  of  lower  trans,  of  Polaris        .         —  1866s 

Error  of  upper  trans,  of  y  Ursae  Majoris  =;      25s 


Sum  of  errors         .         .         .         .         —  1891s 
Then  the  proportion 

1891*  :  653s  =  1000"  :  azimuth  error, 
gives 

azimuth  error  —  345"  =  5'  45''  W. 

6.  An  observer,  at  Boston,  wishing  to  determine  his  merid- 
ian line,  in  the  evening  of  December  17,  1839,  observed  by 
means  of  a  clock  regulated  to  solar  time,  the  superior  transit 
of  «  Cassiopeae  at  6h  4Sm  35s  P.  M.,  and  that  of  Polaris  at 
6h  53"*  15s  P.  M.  What  was  the  azimuth  error  in  the  posi- 
tion of  the  transit  instrument  ? 

Solution,    By  the  Nautical  Almanac, 

R.  A.  of  Polaris         =  I*    2m  26s 
R.  A.  of  a  Cassiopeae '==  0h  31m  28s 

Sideral  Interval  =  0*  30m  58s 


Solar  Interval  —  0*  30™  53s 

Observed  Interval       —  0h    4m  40s 


Error  of  Interval        =  0h  26m  13s  =  1573s. 

Now  Table  C  gives',  for  1000"  of  azimuth  error  and  the  lati- 
tude of  Boston,    since 


§  45.]  THE    MERIDIAN.  221 

Determination  of  the  meridian  line. 

Dec.  of  a  Cassiopeae  =  55°  40' 

Error  of  trans,  of  Polaris  =  1777* 

Error  of  trans,  of  a  Cassiopeae    s=       26* 


DifF.  of  errors  —  1751* 

Then,  the  proportion 

1751s  :  1573s  =  1000"  :  azimuth  error 
gives 

azimuth  error  z*  900"  :  =  V  30"  E. 

7.  Calculate  the  proportional  logarithm  of  0°  2'  33". 

Ans.     1.8487. 

8.  Calculate  the  proportional  logarithm  of  2°  59'  12". 

Ans.     0.0019. 

9.  Calculate  the  corrections  of  tables  A  and  B,  when  the 
latitude  is  54°,  and  the  star's  polar  distance  20°. 

Ans.     Corr.  A  =  125s. 

Corr.  B  =  38'  48". 

10.  Calculate  the  corrections  of  table  C,  when  the  latitude 
is  20°,  and  the  polar  distance  5°. 

Ans.     For  the  upper  transit,  corr.  C  =  091s. 

For  the  lower  transit,  corr.  C  =  737s. 

11.  An  observer  at  Boston,  in  the  year  1840,  wishing  to 
determine  his  meridian  line,  observed  three  successive  transits 
of  Polaris,  by  means  of  a  clock  regulated  to  solar  time.  The 
first  lower  transit  was  observed  at  6h  A.  M.,  the  next  transit  at 

19* 


222  SPHERICAL    ASTRONOMY.  [ctt.   III. 

Determination  of  the  meridian  line. 

7h  2CT  11s  P.  M.,  and  the  second  lower  transit  at  5*  56m4s  A.  M. 
What  was  the  time  of  the  star's  passing  the  meridian  the 
second  morning?  and  what  was  the  azimuth  error  in  the  po- 
sition of  the  instrument  ? 

Ans.     The  time  of  third  merid.  trans,  was  &1  58m  11s  A.  M. 
The  azimuth  error  =  V  8"  W. 


12.  An  observer  at  Boston,  wishing  to  determine  his  me- 
ridian line  by  means  of  a  clock  regulated  to  solar  time,  ob- 
served the  inferior  tfansit  of  Polaris  on  April  4,  1839,  at 
0h  A.  M.,  and  the  superior  transit  of  n  Ursse  Majoris  at  0h  53m 
59*  A.  M.  What  was  the  azimuth  error  in  the  position  of  his 
transit  instrument? 

The  R.  A.  of  Polaris  is  V1  0m  50*,  that  of  n  Ursse  Majoris  is 
13*  4i»  14%  and  the  declination  of  n  Ursae  Majoris  is  50°  7'  N. 
Ans.     The  azimuth  error  =  T  18"  W. 

13.  An  observer  at  Boston,  wishing  to  determine  his  me- 
ridian line,  in  the  evening  of  May  1,  1839,  observed  by  means 
of  a  clock  regulated  to  solar  time,  the  lower  transit  of  Polaris 
at  9*  4971  22-"  P.  M.,  and  that  of  «  Cassiopeae  at  9*  52m  P.  M. 
What  was  the  azimuth  error  of  the  instrument? 

The  R.  A.  of  Polaris         =  t*    0m  56s. 

The  R.  A.  of  «  Cassiopese  ==  0*  31771  22s. 

The  Dec.  of  «  Cassiopese  =  55°  39'  N. 

Ans.     The  azimuth  error  =  18'  23"  W. 


<$>  46.]  LATITUDE.  223 

Latitude  found  by  meridian  altitudes. 


CHAPTER    IV. 

LATITUDE. 

46.  Problem.    To  find  the  latitude  of  a  place. 

Solution.  The  latitude  of  the  place  is  evidently,  from 
(fig.  34.),  equal  to  the  altitude  of  the  pole ;  so  that  this  problem 
is  the  same  as  to  find  the  altitude  of  the  pole,  which  would  be 
done  without  difficulty  if  the  pole  were  a  visible  point  of  the 
celestial  sphere. 

First  Method.    By  Meridian  Altitudes.   [B.  p.  166-175.] 

Observe  the  altitude  of  a  star  at  its  transit  over  the  meridian, 
and  let 

A  z=  the  altitude  of  the  star, 

A1  =z  >fc's  dist.  from  point  of  horizon  below  the  pole  ; 

then,  if  the  notation  of  §  28  is  used,  it  is  evident,  from  (fig.  34.), 
that 

L  =  Af^pp;  (344) 

the  upper  sign  being  used  when  the  transit  is  a  superior  one, 
and  the  lower  sign  when  it  is  an  inferior  one. 

I.  Suppose  the  observed  transit  to  be  a  superior  one  ;  then, 
if  it  passes  upon  the  side  of  the  zenith  opposite  to  the  pole,  we 
have 

A'  =  180°  —  A,    p  =  90°  =p  2>, 


224  SPHERICAL    ASTRONOMY.  [CH.  IV. 

Latitude  found  by  meridian  altitudes, 
arid  (344)  becomes 

L=z90°  —{A±D)  =  (90  —  A)±D  —  z±D  ,     (345) 

the  upper  sign  being  used  when  the  declination  and  latitude 
are  of  the  same  name,  and  the  lower  sign  when  they  are  of 
different  names. 

But  if  the  star  passes  upon  the  same  side  of  the   zenith 
with  the  pole,  we  have 

A'  =  A,    p  =  90°  —  D} 

and  (344)  becomes 

L  —  (A  -f  D)  —  90°  =  D  —  (90°  —  A)  =  D  —  z.    (346) 

II.  If  the  transit  is  an  inferior  one,  we  have 

A'  —  Ay    p  —  90°  —  Dy 

and  (345)  becomes 

L  =x  (A  —  D)  +  90°  =  A  +  (90°  —  D).  (347) 

Equations  (345)  and   (346)  agree  with  the  rule  of  Case  I, 
[B.  p.  166.],  and  (347)  with  Case  II,  [B.  p.  167.] 

III.  If  both  transits  are  observed,  and  if  A*  and  A  are  re- 
ferred to  the  upper  transits,  and 

A  2  =  the  altitude  at  the  lower  transit, 

we  have,  by  (344), 

Lz=z  A  —p 

the  sum  of  which  is 

L  =  i{A'  +  A1);  (348) 


<§>  46.]  LATITUDE.  225 

Latitude  found  by  a  single  altitude. 

so  that  the  latitude  is  determined  in  this  case  without  knowing 
the  star's  declination. 


Second  Method.     By  a  Single  Altitude. 
Observe  the  altitude  and  the  time  of  the  observation. 

I.  If  the  star  is  considerably  distant  from  the  meridian,  we 
have  given  in  the  triangle  PBZ  (fig.  35.),  PB,  BZ,  and 
BPZ  to  find  PZ,  which  may  be  solved  by  Sph.  Trig.  §  59, 
and  gives,  by  the  notation  of  §  28, 

tang.  PC  —  cos.  h  tang,  p  =  ±  cos.  h  cotan.  D     (349) 

cos.  ZC  z=  cos.  PC.  cos.  z  sec.  p 

—  ±  cos.  PC .  cos.  z  cosec.  D,  (350) 

in  which  the  upper  sign  is  used  if  the  declination  and  latitude 
are  of  the  same  name,  otherwise  the  lower  sign. 

90°  —  L  =  PZ  =  PC  ±  ZC 

L  =  90°  —  (PC  do  ZC) ;  (351) 

in  which  both  signs  may  be  used  if  they  give  values  of  L 
contained  between  0°  and  90°,  and  in  this  case  other  data 
must  be  resorted  to,  in  order  to  determine  which  is  the  true 
value  of  L. 

Scholium.  The  problem  is,  by  Sph.  Trig.  §  61,  impossible, 
if  the  altitude  is  greater  than  the  declination,  when  the  hour 
angle  is  more  than  six  hours. 

II.  If  the  latitude  is  known  within  a  few  miles,  it  may  be 
exactly  calculated  by  means  of  (317),  or 

cos.z:=  cos.  [90°— (L+p)]—  2cos.£cos.Z>(sin.p)2.  (352) 


226  SPHERICAL    ASTRONOMY.  [CH.   IV. 

Latitude  found  by  a  single  altitude. 

But  if  A  is  the  star's  observed  altitude,  and^42  its  meridian 
altitude  at  its  upper  transit,  (344)  gives 

Ax  —  L  +  p,  or  =  180°  —  (L+p), 

and  (352)  becomes,  by  transposition, 

sin.  A  2  =  sin.  A  -f-  2  cos.  L  cos.  D  (sin.  J  A)2  ;       (353) 

from  which  the  meridian  altitude  may  be  calculated  by  means 
of  table  XXIII,  as  in  the  Rule.    [B.  p.  200.] 

III.  A  formula  can  also  be  obtained  from  (281),  which  is 
particularly  valuable  when  the  star  is,  as  it  always  should  be 
in  these  observations,  near  the  meridian. 

In  this  case  we  have  in  (281)  applied  to  PBZ 

2sz=90°  —  L-\-p+z  — 180°  —  L  +  p  —  A 

2s—2PZ—L+p  —  A 

=zA1—Aov  =  180°  —  (Ax  +  A)        (354) 

25— 2PBz=  180°  —  L—p  —  A 

—  180°  —  (A1  +  A)ov=A1  —  A;     (355) 

and  if  these  values' are  substituted  in  (281),  after  it  is  squared 
and  freed  from  fractions,  they  give 

(sin.  J  h)2  cos.  L  cos.  Z>:=sin.  J  (^1  x — ^4)  cos.  J  (A  l-\-A)i  (356) 

or 

sin.J(J.1 — ^l)=(sin.J/«)2cos.JLcos.Z>sec.J(-414-^);  (357) 

and  if,  in  the  second  member  of  this  equation,  the  value  of  At 
is  used,  which  is  obtained  from  the  approximate  value  of  the 
latitude,  the  difference  between  the  observed  and  the  meridian 
altitudes  may  be  found  at  once ;  and  this  difference  is  to  be 
added  to  the  observed  altitude  to  obtain  the  meridian  altitude. 


§  46.]  LATITUDE.  227 

Single  altitude  near  the  meridian. 

IV.   If  the  star  is  very  near  the  meridian,  %(A1  —  A)  and 
J  h  will  be  so  small,  that  we  may  put 

mn.i{A1  —  A)_i(Al—A)_j(A'  —  A)     sm.jh_ 

sin.  1"         -         I"         —  1  '"  '     sin.F    —  *    ' 

or  sin.  i  (A1  —  A)  =  i  (A'  —  A)  sin.  \" 

sin.  J  li  —  J  h  sin.  1*  =  l5  h  sin.  1"  ; 

which,  substituted  in  (357),  give,  by  supposing  Ax  equal  to  A 
in  the  second  member,  which  is  very  nearly  the  case, 

A  ,  —  A  =  y>  h2  sin.  1*  cos.  L  cos.  Z>  sec.  Jl  1 .       (358) 

This  value  of  A1  — A  is  proportional  to  h2,  so  that  if  it 
were  calculated  for 

A=  Is, 

any  other  value  might  be  calculated  by  multiplying  byA2.% 
Now  Table  XXXII,  of  the  Navigator,  contains  the  values  of 
A1  —  A  for  all  latitudes  and  for  all  declinations  less  than  24°, 
excepting  a  few  latitudes  in  which  the  meridian  transit  of  the 
observed  body  is  too  near  the  zenith  for  this  observation  to  be 
accurate  ;  and  Table  XXXIII  contains  all  the  values  of  h2, 
where  h  is  less  than  13m. 


V.  If  the  observed  star  is  very  near  the  pole,   we  have  in 
(349) 

tang.  PC  =:  cos.  h  tang,  p  ;  (359) 

so  that  as  p  is  very  small,  PC  must  be  likewise  small,  and  we 
have 

tang.  PC       PC 

cos.  h  = = 

tang,  p  p 

PC  =p  cos.  h;  (360) 


228  SPHERICAL    ASTRONOMY.  [CH.  IV. 

; _ ,  — 

Altitude  of  the  pole  star. 

and,  by  PI.  Trig.  §  22, 

cos.  PC  ==  1,     sin.  D  =  cos.  p  =  1, 

whence,  by  (350),  and  (351), 

cos.  ZC  ==  cos.  z,     ,ZC  =  z9 

L=90o  —  PC—ZC=90°  —  z—PC 

=  A  —  p  cos.  h  ;  (361) 

so  that  p  cos.  7i  may  be  regarded  as  a  correction  to  be  sub- 
tracted from  A  when  it  is  positive,  that  is,  when  the  hour 
angle  is  less  than  6  hours,  or  greater  than  18  hours;  and  it  is 
to  be  added  when  the  hour  angle  is  greater  than  6  hours  and 
less  than  18  hours. 

The  table  [B.  p.  206.]  for  the  pole  star  was  calculated  for 
the  year  1840,  when 

its  R.  A.  =  lh  2m  ;  its  dec.  ==  88°  27'  nearly. 

Third  Method.     By  Circummcridian  Altitudes. 

I.  If  several  altitudes  are  observed  near  the  meridian,  each 
observation  may  be  reduced  separately  by  (357)  and  (358), 
and  the  mean  of  the  resulting  latitudes  is  the  correct  latitude. 

II.  But  if  (358)  is  used,  the  mean  of  the  values  of  Ax — A 
is  evidently  obtained  by  multiplying  the  mean  of  the  values  of 
h2  by  the  constant  factor ;  and  if  to  the  mean  of  the  values 
of  Ax  — A,  the  mean  of  the  values  of  A  is  added,  the  sum 
is  the  mean  of  the  values  of  Alt  whence  precisely  the  same 
mean  of  resulting  latitude  is  obtained  as  by  the  former  method, 
but  with  much  less  calculation. 


<§>  46.]  LATITUDE.  229 

By  circummeridian  altitudes. 

III.  If  the  star  is  changing  its  declination  in  the  course  of 
the  observations,  this  change  may,  in  all  cases  which  can 
occur  if  the  hour  angle  is  small,  be  neglected  in  the  value  of 
cos.  D.  But  the  value  of  Ax  will  not,  in  this  case,  be  at  each 
observation  equal  to  the  meridian  altitude,  but  will  differ  from 
it  by  the  difference  of  the  star's  declination,  Let  the  change 
of  the  star's  declination  in  one  minute  be  denoted  by  <?Z>, 
which  is  positive  when  the  star  is  approaching  the  elevated  pole; 
and  if  h  is  J;he  star's  hour  angle  at  the  time  of  observation, 
which  is  negative  before  the  star  arrives  at  the  meridian  and 
afterwards  positive,  the  whole  change  of  declination  is  hdD, 
so  that  the  correct  meridian  altitude  is 

The  mean  of  the  values  of  the  corrected  meridian  altitude  is, 
therefore,  equal  to  the  mean  of  the  values  of  A1  diminished 
by  the  mean  of  the  values  of  h  $  D  ;  and,  if  H  denotes  the 
mean  of  the  hour  angles  h  (regard  being  had  to  their  signs), 
the  correct  meridian  altitude  is  the  mean  of  the  values  of  Ax 
diminished  by  H$D. 

Fourth  Method.     By  Double  Altitudes. 

I.  Let  two  altitudes  of  a  star,  which  does  not  change  its 
xleclination,  be  observed,  and  the  intervening  time.  Then 
(fig.  39.)  let  ^be  the  zenith,  P  the  pole,  8  and  8'  the  po- 
sitions of  the  star;  join  ZS,  Z8\  PS,  PS',  and  SS'M; 
draw  PI1  to  the  middle  Tof  88',  join  ZT,  and  draw  ZV 
perpendicular  to  PT.     Let 

P=lPS—PS  —  90°  —  D,  SPS'=i  elapsed  time  ==  h 
8T=A  ~8'T,  PT—  90°  —  B 
20 


230  SPHERICAL    ASTRONOMY.  [CH.  IV. 

By  double  altitudes. 

Ax  b£  90°  —  ZS,  A\  =  90°  —  ZS1 
ZTP  =z  T,  ZT=F,  ZV=C 
TV=  Z,  PV=90°  —  E; 

in  which  D  and  B  are  positive,  when  the  latitude  and  decli- 
nation are  of  the  same  name,  but  negative,  if  they  are  of  con- 
trary names  ;  Z  is  positive,  if  the  zenith  is  nearer  the  elevated 
pole  than  the  point  M. 

Now  the  triangle  TPS  gives 

sin.  A  ==  sin.  PS  sin.  SP  T  —  cos.  D  sin.  £  k 

cos.  PS  =:  cos.  P  T  cos.  As  or  sin.  D  =  sin.  B  cos.  A,  (362) 

or   cosec.  A  =  sec.  D  cosec.  £  h  (363) 

cosec,  B  =  cos.  A  cosec.  D.  (364) 

The  triangles  ZTS  and  ZTS'  give 

sin.  A1  =  cos.  Jr  cos.  J.  —  sin.  J1,  sin.  .4  sin.  T,  (365) 

sin.  A ',  saa  cos.  .F  cos.  ^4  -{-  sin.  .F.  sin.  -4  sin.  T,  (366) 

The  sum  and  difference  of  which  is,  by  (36)  and  (37), 

sin.  i{Ax  +  A\)  cos.  £  (A\  —  Ax)  =  cos.  F  cos.  ^4,       (367) 

sin.  J  (■*',  —  4,)  cos.  |  (A|  +  A\)  z^sin.Fsin.^sin.T7.  (368) 

But  triangle  ZTV  gives 

sin.  C  =  sin.  F  sin.  T,  (369) 

cos.  F  —  cos.  C  cos.  ^;  (370) 

which,  substituted  in  (367)  and  (368),  give 

sin. C=  sin.  £(A\  —  A  J  cos.  £  (A  2  +  -4 i)  cosec.  Af       (371) 

sec.JZT=:  cos.  A  cos.Csec.  £  (A  2+ A \ )  cosec.  ^(^i — A  t ).  (372) 


§  46.]  LATITUDE.  231 

By  double  altitudes. 

But  PV=  PT—  TV, 

or  90°  —  E  =  90°  —  B  —  Z 

E  =  B  +  Z.  (373) 

Lastly,  triangle  ZPV  gives 

cos.  PZ  z=  cos.  ZV  cos  PV 

sin.  L  =  cos.  C  sin.  E.  (374) 

Equations  (363,  364,  371  -  374)  correspond  to  the  rule  and 
formula  given  in  the  Navigator.  [B.  p.  180.] 

II.  Another  method  of  calculating  the  values  of  B,  C,  and 
Z  has  been  given,  which  dispenses  with  A  and  one  opening 
of  the  tables,  and  may  therefore  be  preferred  by  some  calcu- 
lators, although  it  requires  one  more  logarithm.  Triangle 
TPS  gives 

tang.  PT  —  cos.  J  h  tang.  PS, 

or  cotan.  B  =  cos.  J  h  cotan.  D.  (375) 

The  substitution  of  (364)  in  (371)  gives 
sin. Cz=  cos. %(A  x-\-A \ )  sin. %(A\ — A  x )  sec.  D  cosec. %h.  (376) 
Triangle  PTS  gives 

cos.  A  ==  sin.  D  cosec.  B ;  (377) 

which,  substituted  in  (372),  gives  (378) 

sec.Z=  cos.Csin.  Z>  cosec.  B  cosec.  ±(A  1-f-^i;)sec.^(^41' — A  ± ). 

Corollary,  The  hour  angle  ZP  T  is  the  mean  between  the 
hour  angles  ZPS  and  ZPS't  and  if  we  put 

ZPTz=H, 


232  SPHERICAL    ASTRONOMY.  [CH.  IV. 

By  double  altitudes.     Douwes's  method. 

the  triangle  ZP  V  gives 

tang.  II  ±=  tang.  C  sec.  E,  (379) 

as  in  B.  p.  181. 

III.  Douwes's  Method.    When  the  latitude  is  known  within 
a  few  miles.     In  this  case  let 

U  =  the  assumed  latitude, 

and  the  triangles  ZSP  and  ZSP  give 

sin.  A  x  =  sin.  L1  sin.  D  -f  cos.  U  cos.  D  cos.  ZPS}     (380) 
sin.  A\  =  sin.  L'  sin.  D  -f-  cos.  L1  cos.  D  cos.  ZPS1 ;  (381 ) 

whence,  and  by  (39), 
sin.  A\ — sin.  A  x  =  cos.  Z/cos.  D  (cos.  ZPS1  —  cos.  ZPS) 

=  2cos.L'cos.Dsm.i(ZPS'-{-ZPS)sm.i(ZPS'--ZPS) 

=  2  cos.  Z/  cos.  Z)  sin.  Z/  sin.  ^-  /a 
2  sin.  Zf=  (sin.  4 'x —  sin.  .4 , )  sec.  L1  sec.  Z>  cosec.  ^  /«,  (382) 
ZPS  =  H  —  l-h]  (383) 

whence  the  hour  angle  ZPS  corresponding  to  the  observation 
at  S'  is  known,  and  the  latitude  may  be  found  by  the  method 
of  a  single  altitude.  The  combination  of  the  formulas  (380, 
381),  and  the  method  of  computing  the  latitude  by  a  single 
altitude,  corresponds  exactly  to  the  rule  given  in  the  Naviga- 
tor. [B.  p.  185.] 

The  log.  cosec.  £  h  is  not  only  given  in  table  XXVII,  but 
also  in  table  XXIII,  where  it  is  called  the  log.  £  elapsed  time 
off*. 


§  46.]  LATITUDE.  233 

Table  XXIII. 

The  value  of 

log.  2  sin.  H  —  5  =  log.  sin.  H  -f-  log.  2 

=  log.  sin.  H  —  ar.  co.  log.  2  -J-  6 

=  log.  sin.  H—  4.69897 

±=  5.30103  =  log.  elapsed  time  of  H     (384) 

is  inserted  in  table  XXIII,  and  is  called  the  log.  middle  time 
of  H.  The  5  is  subtracted  from  log.  2  sin.  if,  on  account  of 
the  different  values  of  the  radius  in  tables  XXIV  and  XXVII. 

Scholium.  When  the  calculated  latitude  differs  much  from 
the  assumed  latitude,  the  calculation  must  be  gone  over  again, 
with  the  calculated  latitude  instead  of  the  assumed  latitude. 
This  labor  may  be  avoided  by  noticing,  in  the  course  of  the 
original  calculation,  the  difference  which  would  arise  from  a 
change  of  10'  in  the  value  of  the  assumed  latitude,  and  calcu- 
lating the  correction  of  the  latitude  by  the  rule  of  double 
position.  The  error  of  the  hypothesis  is  in  each  case  the  ex- 
cess of  the  calculated  above  the  assumed  latitude,  and  the 
proportion  is 

diff.  of  errors  :  diff.  of  hyp.  =  least  error  :  corr.  of  hyp.  (385) 

IV.  If  the  star  has  changed  its  declination  a  little,  during 
the  interval  between  the  observations,  the  second  altitude  will 
correspond  to  a  declination  D',  a  little  different  from  D. 

If  D1  is  put  instead  of  D  in  (381),  and  if  A2  denotes  the 
second  observed  altitude,  A  x  being  retained  to  denote  what 
this  second  altitude  would  have  been,  if  the  declination  had 
remained  unchanged,  (381)  becomes 

sin.  A'2  =  sin.  L'  sin.  D'  +  cos.  L'  cos.  D'  cos.  ZPS'.  (385) 
20* 


234  SPHERICAL   ASTRONOMY.  [cH.  IV. 

Table  XLVI. 

Now,  if  (381)  multiplied  by  cos.  D'  is  subtracted  from  (385) 
multiplied  by  cos.  D,  the  remainder  is 

cos.  Dsin.il  '2— cos.  2>7  sin.il  j  zz:  sin.Z/sin.(Z>'— D).  (386) 

But  if  we  put 

D'  —  D  =  dD,  A'2  —  A'1=dA1  (387) 

we  have,  by  (13)  and  (15), 

cos.  D  h±  cos.  (D  +  3  D)  =  cos.  D  —  sin.  a  2> .  sin.  Z>  (388) 
sin.  ilgzzisin.  (^4^  +  *A  x  )=sin.  A\-\-&\n.  *AX.  cos.il  j,  (389) 

which,  substituted  in  (386),  give 

sin.il'jSin.Dsin.^Z).  -}-cos.  A\  cos.Z?sin.  ^iljZzzsin.Z/sin.  §D 

sin.  (j^  __  SAX    _  sin.  L1  —  sin.  A\  sin.  D  .        . 

sin.  d  D  Z  ~~  TJj   ~~  cos.  A\  cos.  D  *       ' 

and,  by  (34)  and  (35), 

2sin.Ii7  —  cos.(^,1--.D)  +  cos.(^,1  +  J9)^ 
**p ^(A\ -D)+ cos.  (A>1  +  D)  —  D>  (391) 

in  which  D  is  to  be  negative,  when  the  latitude  and  decli- 
nation are  of  contrary  names.  Hence  the  value  of  d  A1  can 
be  computed  by  this  formula,  and  thence 

A\  —  A'2  — 3  A19 

and  in  calculating  t  At,  A'2  may  be  substituted  fov  A\.  Since 
the  value  of  <S  A  x  is  proportional  to  sD,  it  may  be  computed 
for  some  assumed  value  of  8  D,  a  d  arranged  in  a  table  like 
table  XLVI  of  the  Navigator,  and  the  value  of  $A1  can  be 
computed  from  this  table  by  a  simple  proportion.  The  value 
of  A\  is  thus  found;  the  rest  of  the  calculation  can  be  con- 
ducted according  to  the  preceding  methods,  as  in  B.  p.  189. 


§  46.]  LATITUDE.  235 

By  double  altitudes  of  different  stars. 

V.  If  two  stars  are  observed,  whose  declinations  are  quite 
different.  Then,  if  P  (fig.  40)  is  the  pole,  Z  the  zenith, 
S  and  S'  the  places  of  the  star. 

Ax  =  90°  —  ZS  =  the  less  altitude, 

A\  =  90°  —  ZS1  =  the  greater  altitude, 

D  =  90°  —  PS  =  the  declination  of  star  at  S, 

D>  =  90°  —  PS'  —  the  declination  of  star  at  S', 

H  =  SPS'  ±=  hour  angle  =  interv.  of  sideral  time. 

Then,  in  the  triangle  PSS',  PS,  PS',  and  H  are  given  to 
find 

SS'  =  C,  and  S'SP  =  90°  —  F. 

Next,  in  the  triangle  ZSS',  the  three  sides  are  known,  to 
find  the  angle 

ZSS'  ==  z. 

Hence  ZSD  =  90°  —  #  =  90°  —  jP—  Z 

G  =  F+  Z. 

Lastly,  in  the  triangle  ZSP,  ZS,  SP,  and  the  included  angle 
ZSP  are  given  to  find 

ZP  4  90°  —  L. 

This  solution  is  precisely  similar  to  the  Rule  in  B.  p.  193  ; 
and  it  is  easy  to  prove  the  rules  for  the  s  gns  which  are  there 
given. 

VI.  If  the  distance  SS'  were  observed,  the  angles  ZSS' 
and  S'SP  might  be  found  from  the  triangle  ZSS'  and  S'SP, 
in  which  the  sides  are  all  known,  and  the  rest  of  the  calcula- 
tion would  be  as  in  the  last  method,  and  this  method  corre- 
sponds exactly  to  the  Rule  in  B.  p.  197. 


236  SPHERICAL    ASTRONOMY.  [CH.  IV. 

Meridian  altitudes. 


47.    Examples. 

1.  The  correct  meridian  altitude  of  Aldebaran  was  found 
by  observation,  in  the  year  1838,  to  be  55°  45',  when  its  bear- 
ing was  south  ;  what  was  the  latitude  ? 

Solution  The  zenith  distance  =z  34°  15'  N. 

The  declination         =  16°  10'  N. 


The  latitude  =  50°  25'  N. 

2.  The  correct  meridian  altitude  of  Canopus  was  found  by 
observation,  in  the  year  1839,  to  be  16°  25',  when  its  bearing 
was  south;  what  was  the  latitude? 

Solution.  The  zenith  distance  =  73°  35'  N. 

The  declination         z=  52°  36'  S. 


The  latitude  =  20°  59'  N. 

3.  The  correct  meridian  altitude  of  Dubhe  was  found  by 
observation,  in  the  year  1830,  to  be  50°  45',  when  its  bearing 
was  north ;  what  was  the  latitude  ? 

Solution.  The  zenith  distance  =  39°  15'  S. 

The  declination         =  52°  36'  N. 


The  latitude  =  13°  21'  N. 

4.  If  the  correct  meridian  altitude  of  Dubhe,  at  its  greatest 
elevation,  were  found  by  observation,  in  the  year  1830,  to  be 
50°  45',  when  its  bearing  was  south ;  what  would  be  the  lati- 
tude? 


§   47.]  LATITUDE.  237 

Meridian  altitudes. 

Solution.  The  zenith  distance  —  39°  15'  N. 

The  declination         =  52°  36'  N. 


The  latitude  =  91°  51'  N. 

The  problem  is  impossible. 

5.  The  correct  meridian  altitude  of  Dubhe,  at  its  least  ele- 
vation, was  found  by  observation,  in  the  year  1830,  to  be 
50°  45' ;  what  was  the  latitude  ? 

Solution.  The  polar  distance  ±=  37°  24'. 

The  altitude  ±=  50°  45'. 


The  latitude  =  88°  09'  N. 

6.  The  correct  meridian  altitudes  of  Dubhe,  at  its  greatest 
and  least  elevation,  which  were  on  opposite  sides  of  the  zenith, 
were  found  by  observation  to  be  41°  56'  and  53°  16';  what 
was  the  latitude  ? 

Solution.  The  greatest  altitude  —     53°  16'. 

The  least  altitude        =     41°  56'. 


Diff.  of  altitudes  ==     11°  20;. 

180°  —  Diff.  of  altitudes  =  168°  4(K 

Latitude   '  ~     84°  20'  N. 

7.  The  correct  meridian  altitudes  of  a  northern  star,  at  its 
greatest  and  least  altitudes,  which  were  on  the  same  side  of 
the  zenith,  were  found  by  observation  to  be  12°  14'  and  72° 
14' ;  what  was  the  latitude  ? 


238 


A 


SPHERICAL    ASTRONOMY. 


[CH.  IV. 


Single  altitude. 


Solution. 


Greatest  alt.  —  72°  14'. 
Least  alt.      =  12°  14'. 


Sum  of  alts.  =  84°  28'. 
Latitude        =  42°  14'  N. 

8.  In  a  northern  latitude,  the  altitude  of  Aldebaran  was 
found  by  observation,  in  the  year  1839,  to  be  25°  38',  when  its 
hour  angle  was  4A  12™  20s ;  what  was  the  latitude  ? 

Solution.         By  (349,  350,  351), 
h  =  4*  12™  20s      cos.     9.65580 
Z>=]6°11'       cotan.   10.53729     cosec.  10.55484 


90°— PC  =32°  40' 


ZC  —  33°    6' 


cotan.  10.19309 
A  =  25°  38' 


sin.     9.73215 
sin.     9.63610 


cos.    9.92309 


L  =  65°  46'  N. 


9.  In  lat.  65°  40'  N.  nearly,  the  altitude  of  Aldebaran  was 
found  by  observation,  in  the  year  1839,  to  be  25  38',  when  its 
hour  angle  was  4A  12m  20s;  what  was  the  true  latitude? 


Solution.     I. 

65°  40' 

COS. 

9.61494 

16°  11' 

COS. 

9.98244 

Ah  12m  20* 
Nat.  num.     21657 

log.  Ris. 

4.73823 

4.33567 

25°  38' 

Nat.  sine      43261 

49°  31'  N. 

Nat.  cos.      64918 

16°  11' N. 

65°  42'  N.  =  the  latitude. 


$  47.]  LATITUDE.  239 

Single  altitude. 

Had  the   assumed  latitude  been  taken  10'  more,  the  calcu- 
lated latitude  would  have  been  65°  48J'  N.  ;  hence,  by  (385), 

3£  :  1£  —  10'  :  4'  =  corr.  of  second  hypothesis, 
or  the  latitude  —  65°  46'  N.,  as  in  the  preceding  example. 


II.  By  (357), 


£  h  =  2k  6m  10*  2  log.  sin.  9.43720 
L  =  65°  40'  cos.     9.61494 

D  —  16°  IV  cos.     9.98244 


Ax  =  40°  31' 

A  =z  25°  38'  A'  =  25°  38' 


Al—A=U05V    i(Al  +  A)  =  33°    4f         sec.  10.07678 

A1=40°29'    i(At  —  A)—    7°25J'         sin.     9.11136 
corr.  A 1  =  15°  2'  =z  corr.  lat.  =  65°  40'  +  2  =  65°  42'  as  before. 

10.  Calculate  the  variation  of  a  star's  altitude  in  one  minute 
from  the  meridian,  when  the  declination  is  12°  N.  and  the 
latitude  5°  N. 

Solution.     If  A  2  —  A  is  required  in  seconds,  (358)  gives 

A  j  —  A  —  450  sin.  lm  cos.  L  cos.  D  sec.  Ax 
by  450  sin.  lm  —  log.  450  +  log.  sin  lm 

=  2.65321  -f  7.63982       =       0.29303 
L  ±t  12°  cos.     9.99040 

D  =z  5°  cos.     9.99834 


i4i:=830  sec.     0.91411 


^  —  A  =  15".7,  as  in  table  XXXII.      1.19588 


240  SPHERICAL    ASTRONOMY.  [CH.  IV. 


Single 

i  altitude 

near  the  n 

leridian. 

11.  Calculate 
XXXIII. 

the 

tabular 

number 

for    11 

m  48s   in 

table 

Solution. 

Ilm48s  - 

=  708s 

log. 

2.85003 

60s 

log. 

1.77815 

1.07188 
2 

139.2,  as  in  table  XXXIII.        2.14376 

12.  In  lat.  45°  28'  N.  nearly,  the  correct  altitude  of  Alde- 
baran  was  found  by  observation,  in  the  year  1839,  to  be 
60°  40'  20",  when  its  hour  angle  was  7m  17s.  What  was  the 
true  latitude,  if  the  declination  of  Aldebaran  was  16°  11' 
9".2  N.  ? 

Solution. 


From  Table  XXXII 

2".7 

From! 

'able  XXXIII 

2  23".  1  = 

53 

143".  1 

60° 

40'  20" 

Third  alt. 

=  60° 

42'  43".  1 

Dec. 

—  16° 

11'    9 '.2 

Lat. 

=  45° 

28'26".l  N. 

13.  In  lat.  40°  N.  nearly,  the  sum  of  ten  correct  central 
altitudes  of  the  sun,  when  its  declination  was  20°  S.  were 
300°  20'.  The  hour  angles  of  these  observations  were 
4"  15s,  3",  2W  6%  lm8s,  30s,  50s,  lm  12s,  2m  15s,  3"  10s,  4OT  25s. 
What  is  the  true  latitude,  if  the  change  of  declination  is  neg- 
lected ? 


§  47.]  LATITUDE.  241 

Latitude  by  circummeridian  altitudes. 

Solution.     The  numbers  of  Table  XXXIII  are 
4m  15s       gives      18.1 


3 

0 

9.0 

2 

6 

4.4 

1 

8 

1.3 

0 

30 

0.2 

0 

50 

0.7 

1 

12 

1.4 

2 

15 

5.1 

3 

10 

10.0 

•        4 

25 

19.5 

Sum    == 

69.7 

Mean  = 

6.97 

Table  XXXII  gives 

1".6 

11" 

Mean  of  observations  =r  30°    0'  40" 


Merid.  alt.     =  30°    0'  51" 
Dec.  =  20°  S. 


Lat.  z=39°59'  9"  N. 

14,  AtGottingen,  in  lat.  51°  32'  N.  nearly,  the  correct  central 
altitudes  of  the  sun  on  the  11th  of  March,  1794,  were  by 
observation 

21 


Ml  SPHKRICJLL  *T.  [CH,  IT. 

Laftaafe  fey  * 


«*ew*s 

_9M1* 

*i 

19 

—     1 

h 

—  5 

16 

—  1 

m 

—  2 

47 

0 

19 

-2 

5 

3 

9 

4 

■ 

»6 

S 

declination  was  ^  30  a*   S.,  and  it 
at  the  rate  of  0.9$  in  a  minote.  What  is  the 


of  the  ahkodes  is  $4*  56  »  .5  : 
of  Table  XXXm  is 
by  1  ,5  from  Ta- 


The  Mean  of  the  boor  angles  is,  regard- 
ing their  signs,  — 1»  50%  which,  multiplied 
by  0  J98,  gwes  by  (363),  for  the  correction 

of  :ne  MrifiMi  nttSE  I     5 


The  meridian  altitude  =  S4    57   II 
The  declination  —  K       S 


Tiff  hi ton*  =  51?»    4   TN, 


^  47%]  ok.  Mtt 


Hide  by  pole  star. 


i  agrees  exactly  with  the  calculations  of  Littrow  in  his 

15.  Calculate  the  correction  for  the  altitude  of  the  pole  star 
ben  the  right  ascension  of  the  zenith  is  &*  ?"*. 

>  tot©*  By  (301), 

*  iw>w=i^  sec.  o.oi:: 

j>  =  1°  33*  Prop.  log.  O.SSoS 


Corr.  alt.  =  1  \\  the  table,     Prop.  log.  0.3045 

u>  Wlu  n  the  ti^ht  ascension  of  the  zenith  was  7*9^*,  the 
altitude  of  the  pole  star  was  obsenod  at  Newburvport  to  be 
4<<T  44'.     What  is  the  latitude  of  Newburvport  ? 

Solution.        The  correction  of  table  ==     0°    & 
Altitude         •  =  43°  44 


Latitude         .        .        s=  49     LI 
IT.  Calculate  the  log,  elapsed  time  and  log.  middle  time  of 

,  win  foi  a  ;  ■  u>. 

flbb/toN.     By  Table  XXVII  and  (384), 

3*  7*  10*    cosec.  0.13735  =  log.  elapsed  time 
5.30103 


5  16968  =  log.  mid,  time. 


244  SPHERICAL    ASTRONOMY.  [CH.  IV. 

Variation  of  star's  altitude. 

18.  Calculate  the  variation  of  the  altitude  of  a  star  arising 
from  the  change  of  100  seconds  in  the  declination,  when  the 
latitude  is  40°,  the  declination  10°,  and  the  altitude  30°. 

Solution.     By  (391), 

L'  z=z  40°,  2  X  Nat.  sin.  1.2856  1.2856 

A\  —  D  =z  20°  Nat.  cos.  0.9397    —  0.9397  0,9397 

A\  +  D=:  40°  Nat.  cos.  0.7660         0.7660  —0.7660 


1.7057  1.1119         1.4593 

1.7057         (ar.  co.)      9.7681         9.7681 
100"  X  1.1119  2.0661 

100"  X  1.4593  2.1641 


65"  =;  var.  when  D  is  -f,       1.8142 


86"  =z  var.  when  D  is  — ,  1.9322 

19  The  moon's  correct  central  altitude  was  found,  by  ob- 
servation, to  be  53°  43',  when  her  declination  was  14°  16'  N. 
After  an  interval,  in  which  the  hour  angle  was  lh  44m  15%  her 
correct  central  altitude  was  42°  29',  and  her  declination  13° 
52'  N.  The  latitude  was  48°  50'  N.  nearly ;  what  was  it 
exactly  ? 

Solution.     Table  XLVI  gives,  for  the  second  alt.      83" 
Whole  change  of  declination  24' 

Correction  of  second  altitude  20' 

Corrected  second  alt.  =  42°  49',     dec.  =  14°  16'  N. 


§  47.]  LATITUDE.  245 

Latitude  by  double  altitudes. 

I.    By  Bowditch's  first  method. 
lh  Um  15*  cosec.  0.64675 
14°  16'         sec.  0.01360         cosec.  0.60830 


A  cosec,  0.66035  cos.  9.98936      cos.  9.98936 


B  =  14°  38'  N.  cosec.  0.59766 

cos.  9.82326  £  sum  alts.=z48°  16'  cosec.  0.12712 
sin.    8.97762  %  diff.  alts.=  5°  27'      sec.  0.00197 


C  sin.   9.46123  cos.  9.98103       cos.  9.98103 


Z=37°19/N.  sec.  0.09948 


E  =  51°  57'  N.  sin.  9.89624 


Latitude  m  48°  55J'N.  sin.  9.87727 

II.    By  the  method  (375  -  378). 
lh  Mm  15s  cos.  9.98852  cosec.  0.64675 

14°  16'  cotan.  0.59469      sec.  0.01360      sin.  9.39170 


B  55  14°  38'  N.    cotan.  0.58321  cosec.  0.59753 

J  sum  alts.  =  48°  16'     cos.  9.82326  cosec.  0.12712 
J  diff.  alts.=  5°  27'     sin.  8.97762     sec.  0.00197 


C  cos.  9.98103      sin.  9.46123      cos.  9.98103 


Z=37°18'N.  sec.    .09935 


jEJ=z51°56'N.        sin.  9.89614 


Lat  =  48°54£'N.   sin.  9.87717 
21* 


246  SPHERICAL   ASTRONOMY.  [CH. 

Latitude  by  double  altitudes. 

III.    By  Douwes's  method. 

48°  50'     sec.  0.18161 
53°  43'     N.  sin.  80610  14°  46'     sec.  0.01360 


42°  49'     N.  sin.  67965  log.  ratio  0.19521 


12645  log.  4.10192 

I  (lh  44w  15')  =     52w    7£s     log.  el.  time  0.64689 


lh  Um  18*     log.  mid.  time  4.94402 


52m  ll£*  log.  ris.  3.41152 

log.  ratio  0.19521 


1645 

log.  3.21631 

80610 

34°  40'  N. 

N.  cos.  82255 

14°16/N. 

Lat.  =  48°  56' N. 

Had  the  latitude  been  supposed  10'  greater,  the  calculated 
latitude  would  have  been  48°  55'  N. 


§  47.]  LATITUDE.  247 

Latitude  by  double  altitudes. 

IV.    By  Bowditch's  fourth  method. 

lh  Um  15s  sec.  0.04657  tan.  9.68938 

14°  16'  N.  tan.  9.40531       sin.  9.39170 


A  =  15°  48'  S.  tan.  9.45188  cosec.  0.56485     cos.  9.98326 

13°52  N. 
B  —    1°  56'  S.  cos.  9.99975  cosec.  1.47190 


C      25°  16'  cosec.  0.36961      cos.  9.95630 


F~    4°    6'N.  cotan.  1.14454 

53°  43'  Z="51°38'N. 


G  —  55°44'N.      sin.  9.91720 
42°  29'     sec.  0.13225       sin.  9.82955  cotan.  0.03820 


£  sum  =60°  44'     cos.  9.68920   J  sec.  0.12936     tan.  9.95540 
Rem.  =   7°    V     sin.  9.08692  K  sin.  9.91823  J~  42°    4'N. 


2)  19.27798  lat.  sin.  9.87714  13°52'N. 


£  Z  =i  25°  49'  N.  sin.  9.63899  lat.  =  48°  54'  N.  K=  55°  56'  N. 

19.  The  correct  meridian  altitude  of  Aldebaran  was,  by 
observation,  56°  25'  40"  bearing  south,  and  its  declination  at 
the  time  of  the  observation  was  16°  S'  44"  N.  ;  what  was  the 
latitude  ? 

Arts.     49°  43'  4"  N. 


248  SPHERICAL    ASTRONOMY.  [CH.  IV. 

Latitude  by  meridian  altitudes. 

20.  The  correct  meridian  altitude  of  Sirius  was  70°  59'  33" 
bearing  north,  and  its  declination  16°  28'  9"  S.  ;  what  was  the 
latitude  ? 

Ans.     26°  28'  36"  S. 


21.  The  meridian  altitude  of  the  sun's  centre  was  25°  38'  30" 
bearing  south,  and  its  declination  22°  18'  14"  S. ;  what  was 
the  latitude  ? 

Ans.     42°  3'  16"  N. 

22.  The  meridian  altitude  of  the  planet  Jupiter  was  50° 
20'  8"  bearing  south,  and  its  declination  18°  47'  37"  N. ;  what 
was  the  latitude? 

Ans.     58°  27'  29"  N. 

23.  The  altitude  of  the  pole  star  was  30°  1'  30"  below  the 
pole,  and  its  polar  distance  1°  38'  2"  ;  what  was  the  latitude? 

Ans.     31°  39'  32"  N. 

24.  The  altitude  of  Capella  on  the  meridian  below  the  pole 
was  9°  52'  42",  and  its  polar  distance  44°  11'  33" ;  what  was 
the  latitude? 

Ans.     54°  4'  15"  N. 

25.  The  meridian  altitude  of  the  sun's  centre  was  7°  9'  11" 
below  the  pole,  and  its  declination  23°  8'  17"  N. ;  what  was 
the  latitude  ? 

Ans.     74°  0'  54"  N. 

26.  The  two  meridian  altitudes  of  a  northern  circumpolar 
star  were  61°  49'  13"  and  47°  34'  27"  ;  what  was  the  latitude  ? 

Ans.     54°  36'  50"  N. 


§  47.]  LATITUDE.  249 

Latitude  by  single  altitudes. 

27.  In  a  northern  latitude,  the  altitude  of  the  sun's  centre 
was  54°  9',  when  its  hour  angle  was  32m  40s,  and  its  declina- 
tion 11°  17'  N. ;  what  was  the  latitude? 

Ans.     4G°  27'  N. 

28.  In  latitude  49°  17'  N.  nearly,  the  altitude  of  the  sun's 
centre  was  14°  15',  when  its  hour  angle  was  lh  40™,  and  its 
declination  23°  28'  S.;  what  was  the  true  latitude? 

Ans.     48°  55'  N. 

29.  Calculate  the  variation  of  a  star's  altitude  in  one  minute 
from  the  meridian,  when  the  declination  is  3°  and  the  lati- 
tude 7°.       * 

Ans.     It  is  27".9  when  the  dec.  and  lat.  are  of  the  same 
name,  and  11".2  when  they  are  of  contrary  names. 

30.  Calculate   the   tabular   number   for   I2m  59'  in  Table 

XXXIII. 

Ans.     168.6. 

31.  In  lat.  50°  30'  N.  nearly,  the  altitude  of  Sirius  was 
22°  59'  36",  when  its  hour  angle  was  4m  15%  and  its  declina- 
tion 16°  29;  ll"  S.  ;  what  was  the  true  latitude? 

Ans.     50°  30'  49"  N. 

32.  In  lat.  20°  27'  N.  nearly,  the  sum  of  seven  altitudes  of 
Sirius  was  371°  21';  the  hour  angles  of  the  observations 
were  7m,  5m  3s,  2m  12*,  9%  3™,  4™  6%  8m  13';  what  was  the 
true  latitude,  if  the  declination  of  Sirius  was  16°  29'  30"  ? 

Ans.     20°  26'  18"  N. 


250  SPHERICAL    ASTRONOMY.  [CH.   IV. 

Latitude  by  circummeridian  altitudes. 

33.  In  lat.  50°  N.  nearly,  the  sum  of  twelve  central  alti- 
tudes of  the  moon  was  590° ;  the  hour  angles  of  the  observa- 
tions were  —  9m  3s,  — -7™  40%  —  6W  12%  —  5*30%  —  3™  2% 
—  1*,  —  12%  50%  \m  59%  4°%  lm  30%  10™  ;  the  moon's  me- 
ridian declination  was  19°  10'  58".4  N.,  and  her  change  of  de- 
clination for  one  minute  13".875  ;  what  was  the  true  latitude  ? 

Ans.     59°  50'  2".3  N. 

34.  Calculate  the  correction  for  the  altitude  of  the  pole  star 
[B.  p.  206.],  when  the  right  ascension  of  the  zenith  is  9h  7m. 

Ans.     48'. 


35.  The  altitude  of  the  pole  star  was  25°  9%  when  the  right 
ascension  of  the  zenith  was  21°  47';  what  was  the  latitude? 

Ans.     24°  8'  N. 

36.  Calculate  the  log.  elapsed  time  and  log.  middle  time  of 
Table  XXIII  for  &  49-  50s. 

Ans.     Log.  elapsed  time  =  0.00001 
Log.  middle  time  —  5.30102 

37.  Calculate  the  variation  of  the  altitude  of  a  star  arising 
from  the  change  of  100  seconds  in  declination,  when  the  lati- 
tude is  60°,  the  declination  20°,  the  altitude  30°,  and  the 
declination  and  latitude  of  the  same  name. 

Ans.     85". 

38.  Calculate  the  variation  of  the  altitude  of  a  star  arising 
from  the  change  of  100  seconds  in  declination,  when  the  lati- 
tude is  50°,  the  declination  24°,  and  the  altitude  20°. 

Ans.     It  is  73"  when  the  lat.  and  dec.  are  of  the  same 
name,  and  105"  when  they  are  of  contrary  names. 


<§>  47.]  LATITUDE.  251 

Latitude  by  double  altitudes. 

39.  The  sun's  correct  central  altitudes  were  found  by  ob- 
servation to  be  30°  13'  and  50°  4';  his  declination  was  20° 
7'N.,  and  the  interval  of  sideral  time  between  the  observations 
was  2h  55m  32s ;  the   assumed  latitude  was  56°  29'  N.  ;  what 

was  the  true  latitude? 

Ans.     56°47'N. 

40.  The  sun's  correct  central  altitude  was  41°  33'  12",  his 
declination  14°  N.  ;  after  an  interval  of  lh  30m,  his  correct 
central  altitude  was  50°  V  12",  and  declination  13°  58'  38"  N.; 
the  assumed  latitude  was  52°  5'  N. ;  what  was  the  true  lati- 
tude ? 

Ans.     52°  5'  N. 

41.  The  moon's  correct  central  altitude  was  55°  38%  her 
declination  0°  20'  S. ;  after  an  interval  in  which  the  hour  angle 
was  5*30*  49%  her  correct  central  altitude  was  29°  37%  and  her 
declination  1°  10'  N. ;  the  assumed  latitude  was  23°  25'  S.  ; 
what  was  the  true  latitude  ? 

Ans.     23°  24'  S. 

42.  The  sun's  correct  central  altitude  was  16°  6%  his  decli- 
nation 8°  18'  N. ;  after  an  interval  in  which  the  hour  angle 
was  3\  his  correct  central  altitude  was  42°  14'  9",  and  his 
declination  8°  15'  N. ;  the  assumed  latitude  was  49°  N. ;  what 
was  the  true  latitude? 

Ans.     48°  50'  N. 

43.  The  moon's  correct  central  altitude  was  35°  21%  and  her 
declination  5°  31'  6"  S. ;  after  an  interval  in  which  the  hour 
angle  was  2h  20™,  her  correct  central  altitude  was  70°  1%  and 
her  declination  5°  28' 54"  S. ;  the  assumed  latitude  was  1°  30'  N.; 
what  was  the  true  latitude  ? 

A?is.     1°  29'  N. 


252  SPHERICAL    ASTRONOMY.  [CH.  IV. 

Corrections  for  the  pole  star  in  the  Nautical  Almanac. 


44.  The  altitude  of  Capella  was  60°  45'  36",  and  her  decli- 
nation 45°  48'  21"  N. ;  at  the  same  instant,  the  altitude  of 
Sirius  was  17°  54'  12",  and  his  declination  16°  28'  40"  S. ; 
the  hour  angle  was  [k  33™  45?,  and  the  latitude  was  about 
53°  15'  N. ;  what  was  the  true  latitude? 

Ans.     53°  19'  N. 

45.  The  altitude  of  «  Bootis  was  50°  3'  39",  and  its  decli- 
nation 20°  10'  56"  N. ;  the  altitude  of  «  Aquilje  was  33°  33', 
and  its  declination  8°  22'  35"  N. ;  the  hour  angle  of  the  ob- 
servations was  5^  5m  5hs,  and  the  assumed  latitude  38°  27'  N.  ; 
what  was  the  true  latitude? 

Ans.     38°  28'  N. 

46.  The  distance  of  the  centres  of  the  sun  and  moon  was 
found,  by  observation,  to  be  75° ;  the  sun's  central  altitude 
was  37°  40' ;  the  moon's  central  altitude  was  55°  20' ;  the  sun's 
declination  was  0°  17'  S. ;  the  moon's  declination  was  0°  36'  N. ; 
what  was  the  latitude,  supposing  it  to  be  north  ? 

Ans.     23°  24'  N. 

48.  The  method  of  determining  the  latitude  by  means  of 
the  pole  star  is  so  accurate  in  practice,  that  tables  are  given 
in  the  Nautical  Almanac  for  correcting  the  observed  altitude 
for  differences  of  latitude,  and  changes  in  the  right  ascension 
and  declination  of  the  star. 

The  first  correction  of  the  Nautical  Almanac  corresponds 
to  that  of  the  Navigator,  and  is  calculated  by  (361)  for  R.  A. 
of  Polaris  =  1A  1»  48*.2.  (392) 

Dec.  of  Polaris  —  88°  26'  54"  =  2>,  (393) 


§  48.]  LATITUDE.  253 

Corrections  of  the  Nautical  Almanac  for  Polaris. 

which  gives 

p  =  1°  33'  6"  ss  5586"  (394) 

log.  p  =  3.74710  (395) 

A  =  R.  A.  of  zenith  —  It  lm  48*.2.     (396) 

The  second  correction  of  the  Nautical  Almanac  depends 
upon  the  latitude,  and  would  vanish,  if  in  (350)  the  values  of 
p  and  PC  were  so  small  that  we  could  put 

sin.  D  =  cos.  p -ss  cos.  PC  ss  1. 

Now  equation  (350)  is  equivalent  to  the  proportion 

cos.  PC  :  cos.  p  ss  cos.  ZC :  sin.  A, 

but,  by  (360), 

ZC  ss  90°  —  L  —  PC=  90°  —  L  —  p  cos.  *; 

whence 

cos.  PC :  cos.  p  ss  sin.  (L  4-  p  cos.  A)  :  sin.  ^4. 

Hence,  by  the  theory  of  proportions, 

cos.  PC  —  cos.  p        sin.  ( L  -f-  /?  cos.  A)  —  sin.  -4     /QQ~X 
cos.  jPC-j-  cos.p        sin.  (jL-f-p  cos.  A)  -}-  sin.  .4' 

and,  by  (40),  (41),  and  (360), 

tang.  [Jp  (1  —cos.  A)]  .  tang.  [Jp  (1  +  cos.  A)]  s= 

tang.£(Z,+p  cos.  A  —  it)  m 

tang.  J  (X, ■  +  p  cos.  A  +  ,4) '  K       ' 

or,  since  p  and  L  -f-  p  cos.  A  —  .4  are  very  small,  we  may  put 

tang.  [Jp  (1  —  cos.  A)]  ss  £p  (1  —  cos.  A)  tang.  \" 

tang.  [Jp  (1  -|-cos.  A)]  =z  \p  (1  +  cos.  A)  tang.  \" 

22 


254  SPHERICAL    ASTRONOMY.  [CH.   IV. 

Corrections  of  the  Nautical  Almanac  for  Polaris. 

tang.  £  (L-\-p  cos.  h  —  A)  =  J  (L-\-  pcos.h —  A)tang.  \" 
tan.  J  {L-\-p  cos.7i-\-A)—t&n.[L-\-p  cos.Ji — %(L-\-pcos.h — A)] 

—  tang.  (L  -\-p  cos.  h)y 
which,  substituted  in  (390),  give 

\p2  (1  —  cos.2  A)tang.(Z/~|-pcos.7f)  tang.l"— JL-{-pcos.  h — A 
or 

L=p  cos.  h -{-A  +  Jjp2  sin.2  h  tang.( L-{-p  cos.  h)  tang,  t"  (399) 
so  that 

£  p2  sin.2  A  tang.  (L  +  p  cos.  A)  tang.  1 "  (400) 

or  J  p2  sin.2  A  tang  L  tang.  1" 

is  the  correction  depending  upon  the  latitude,  and  in  calculat" 
ing  it  we  have 

log.  (£  p2  tang.  1")  s  7.49420  +  9.69897  +  4.68558 

==  1.87875  (401) 

and  7i  is  the  same  as  in  (396). 

The  third  correction  of  the  Nautical  Almanac  is  the  change 
in  the  value  of  the  first  correction  arising  from  the  changes  in 
the  declination  and  right  ascension  of  the  star.  Thus  if  the 
declination  is  greater  than  that  of  (393)  by  <*  Z>,  the  value  of 
p  must  be  less  by  d  D,  and  the  correction  — p  cos.  h  is  in- 
creased by 

S  D  cos.  h.  (402) 

Again,  if  the  right  ascension  is  greater  than  that  of  (392) 
by  $  R,  the  value  of  h  must  be  less  by  J  R,  and  the  value  of 
—  p  cos.  h  is  increased  by 

— p  [cos.  (h  —  dR)  —  cos.  h]i  (403) 

which,  by  (15),  is  equal  to 


§  49.]  LATITUDE.  255 

Corrections  of  the  Nautical  Almanac  for  Polaris. 

—  p  sin.  $  R  sin.  h  =  — p  d  R  sin.  Is  sin.  h 

—  5580  X  0.000075  X  dR  s\n.h=z0"AdRsm.h,  (404) 

and  the  whole  change  is  the  sum  of  (402)  and  (404).  The 
values  of  (402)  and  (404)  are  easily  obtained  from  the  tables 
of  difference  of  latitude  and  departure.  We  may  neglect  the 
lm  48*.2  in  the  value  of  h  (396),  when  we  calculate  these  cor- 
rections, and  take 

h  —  R.  A.  of  zenith  —  1\  (405) 

The  third   correction  is  sometimes  positive  and  sometimes 
negative,  but  always  less  than  1',  so  that 

V  -f-  the  third  correction 

is  always  positive ;  and  this  is  the  given  sum  in  the  Nautical 
Almanac  ;  that  is,  the  third  correction  is  given  1'  greater  than 
its  real  value,  so  that  it  may  always  be  positive.  The  latitude, 
obtained  by  means  of  the  table  of  the  Nautical  Almanac, 
would  then  be  V  greater  than  its  true  value,  if  V  were  not 
subtracted  agreeably  to  the  rule  given  in  the  Almanac. 


49.    Examples. 

1.  Calculate  the  first  correction  of  the  Nautical  Almanac 
when  the  R.  A.  of  the  zenith  =d  4*  20w. 

Solution.  log.  p  =      3.74710 

L  =  ±h  20"  —  lh  lm  48*.2  =  3h  18-  11\8   cos.  9.81219 


1st  corr.  ==  3624"  —  1°  0'  24"  3.55929 

2.  Calculate  the  second  correction  of  the  Nautical  Almanac 
when  R.  A.  of  the  zenith  is  lh  30m,  and  the  latitude  50°. 


256  SPHERICAL   ASTRONOMY.  [CH.  IV. 

Corrections  of  the  Nautical  Almanac  for  Polaris. 

Solution.  log.  (£p2  tang.  1")     2=     1.87875 

L  =  6h  28m  11».8  sin.2     9.99340 

50°  tang.     0.07619 


2d  corr.  =  89"  =  1'  29"  1.94834 

3.  Calculate  the  third  correction  of  the  Nautical  Almanac 
for  Dec.  31,  1839,  when  R.  A.  of  zenith  ==  14\ 

Solution.       h=Uh—lh=i  13A  p=  12*  +  lh  =  180°  +  15°. 
Dec.  of  Polaris  =  88°  27'  46".6,      R.  A.  —  lh  lm  59s .47 

D  =s  88°  26'  54"  lh  lm  48'.2 


*D=:  52".6  <*22  z=z  11*.27 

0'A$R  =   4".5 

By  Table  II.,  omitting  the  tenths  of  seconds  in  the  result, 

V  +  3d  corr.  —  V  —  50".8  +  1".2  —  1'  —  50"  —  10". 

4.  The  correct  altitude  of  Polaris  on  June  25,  1839,  was 
47°  28'  35",  when  the  Right  ascension  of  the  zenith  was  6h 
18m  30s;  what  was  the  latitude? 

Solution.  Cor.  Alt.     =     47°  28'  35" 

First  corr.     =   —     13' 27" 


A  +  First  corr.     rs     47°  15'    8" 
Second  corr.     =  1'  28" 

Third  corr.       =  1'    6"  —  I1 


Lat.  sr     47°  16'  42"  N. 


<§>  50.]  LATITUDE.  257 

Observer  in  motion. 

5.  Calculate  the  three  corrections  of  the  Nautical  Almanac 
for  Sept.  1,  1839,  and  latitude  70°,  when  the  R.  A.  of  zenith 
is  8h.     At  this  time,  we  have 

Dec.  of  Polaris  =  88°  27'  8".4,     its  R.  A.  =  t*  2m  21*.32. 
Ans.     The  first  correction       =  23'  23" 
The  second  correction  =z    3'  15" 
1'  -f  The  third  correction     =        43". 

6.  The  correct  altitude  of  Polaris  on  March  6,  1839,  when 
the  R.  A.  of  the  zenith  was  6*  39™  24s,  was  46°  17'  28";  find 
the  latitude.  The  following  is  an  extract  from  the  tables  of 
the  Nautical  Almanac  sufficient  for  the  present  example. 


1st  corr. 

2d  corr. 

Lat.  =s  45° 

Lat.z=50{ 

6h  30m 

—  12'  53" 

V  14" 

r  28" 

6*40™ 

—    8' 51" 

1'  16" 

V  30" 

Third  correction  -{-  1/ 
March  1,  April  1, 

6*  V  27"  1'  27" 

8h  V  13"  V  19" 

Ans.    46°  10y  3"  N. 

50.  The  observer  has  been  supposed  stationary  in  the  pre- 
ceding observations,  but  if  he  is  in  motion  his  second  altitude 
will  differ  from  the  altitude  for  this  time  at  the  first  station  by  the 
number  of  minutes  by  which  the  observer  has  approached  the 
star  or  receded  from  it ;  so  that  the  correction  arising  from 
this  change  of  place  is  obviously  computed  by  the  method  in 
[B.  p.  183.] 

22* 


258  SPHERICAL    ASTRONOMY.  [cH.  IV. 

Greatest  altitude  of  a  star  in  motion. 

51.  In  observing  the  meridian  altitude  of  a  star,  the  position 
of  the  meridian  has  been  supposed  to  be  known;  but  if  it 
were  not  known,  the  meridian  altitude  can  be  distinguished 
from  any  other  altitude  from  the  fact  that  it  is  the  greatest  or 
the  least  altitude;  so  that  it  is  only  necessary  to  observe  the 
greatest  or  the  least  altitude  of  the  star. 

52.  But  if  the  star  changes  its  declination,  the  greatest  alti- 
tude ceases  to  be  the  meridian  altitude.  Let  h  denote  the 
hour  angle  of  the  star  at  the  time  of  observation.  Then  if 
the  star  did  not  change  its  declination,  and  if  B  were  the 
number  of  seconds  given  by  Table  XXXII  ibr  the  diminution 
of  altitude  in  one  minute  from  the  meridian  passage,  h2  B 
would  be  the  diminution  of  altitude  in  h  minutes.  But,  since 
h  is  small,  the  altitude,  at  this  time,  is  increased  by  the  change 
of  declination  ;  so  that  if  A  is  the  number  of  minutes  by 
which  the  star  changes  its  declination  in  one  hour,  that  is,  the 
number  of  seconds  by  which  it  changes  its  declination  in  one 
minute,  h  A  will  be  the  increase  of  altitude  in  (he  time  h,  so 
that  the  altitude  at  the  time  h  exceeds  the  meridian  altitude  by 

h  A  —  A2  B.  (406) 

If,  then,  h  denotes  the  time  of  the  greatest  altitude,  and 
h-\-dfi  a  time  which  differs  very  slightly  from  the  greatest 
altitude ;  the  greatest  altitude  exceeds  the  altitude  at  the  time 
h-\-dh  by  the  quantity 

(h  A  —  A*  B)  —  [(/*  +  S  h)  A  —  (h  +  S  ft)*  B] 

=zdh[(—A+2Bh)-\-Bdh],  (407) 

and  3  h  can  be  supposed  so  small  that  B$7i  may  be  insensible, 
and  (407)  becomes 

dh(—A  +  2Bh).  (408) 


$  54.]  LATITUDE.  259 

Greatest  altitude  of  a  star  in  motion. 

Now  —  A  -f-  2  B  h  cannot  be  negative,  because  h  is  sup- 
posed to  correspond  to  the  greatest  altitude,  and  cannot  be 
less   than    the    altitude    at  the   time  li  -\-  §  h.     Neither  can 

—  A-{-2Bhbe  positive,  for  the  altitude   at  the  time  h  ex- 
ceeds that  at  the  time  h  —  8  h  by  the  quantity 

_*/,(_  A  +  ZBh), 

which,  in  this  case,  would  be  negative,  and  the  altitude  at  the 
time  h  —  d  li  would  exceed  the  greatest  altitude.     Since,  then, 

—  A  -\-  2  B  h  can  neither  be  greater  nor  less  than  zero,  we 
must  have 

—  A  +  2Bh=zO 

h  =  wm  (409) 


and  this  value  of  A,  substituted  in  (406),  gives 
A*  A^  A* 


2B         42*  4B 


(410) 


for  the  excess  of  the  greatest  altitude  above  the  meridian  alti- 
tude. 

f 

53.  If  the  observer  were  not  at  rest,  his  change  of  latitude 
will  affect  his  observed  greatest  altitude  in  the  same  way  in 
which  it  would  be  affected  by  an  equal  change  in  the  declina- 
tion of  the  star ;  so  that  the  calculation  of  the  correction  on 
this  account  may  be  made  by  means  of  (409)  and  (410)  pre- 
cisely as  in  [B.  p.  169.] 


54.    E 


XAMPLES. 


1.  An  observer  sailing  N.N.W.  9  miles  per  hour  found,  by 
obserration,  the  greatest  central  altitude  of  the  moon,  bearing 


60  SPHERICAL    ASTRONOMY.  [CH.  IV. 

Latitude  by  greatest  altitude. 

south,  to  be  54°  18/ ;  what  was  the  latitude,  if  the  moon's 
declination  was  6°  30'  S.,  and  her  increase  of  declination  per 
hour  16/.52? 

Solution,  J>  's  zenith  dist.  =.  35°  42'  N. 

3>'s  dec.  =     6°30'S. 


Approx.  lat.         —  29°  12'  N. 
J)  's  increase  of  dec.  per  hour  ==  16'.52  S. 
Ship's  change  of  lat.  =£=     8.3 

A     =  24.82,  A*  =  616.0 

By  Table  XXXII           B     —  2.9,  4  B  =     11.6 

Corr.  of  gr.  alt.  =  corr.  of  lat.  z=  52"  =  1'  nearly 

Lat.  =  29°  12'  +  1'  =  29°  13'  N. 

2.  An  observer  sailing  south  12J  miles  per  hour  found,  by 
observation,  the  greatest  central  altitude  of  the  moon  bearing 
south,  to  be  25°  15';  what  was  the  latitude,  if  the  moon's 
declination  was  1°  12'  N.,  and  her  increase  of  declination  per 
hour  18'.5  ! 

/  Ans.     66°  1'  N. 


§  58.]  THE    ECLIPTIC.  26  L 

Obliquity.  Equinoxes.  Signs. 


CHAPTER   V. 

THE    ECLIPTIC. 

55.  The  careful  observation  of  the  sun's  motion 
shows  this  body  to  move  nearly  in  the  circumference 
of  a  great  circle.  This  great  circle  is  called  the  eclip- 
tic,   [B.  p.  48.] 

56.  The  angle  which  the  ecliptic  makes  with  the 
equator  is  called  the  obliquity  of  the  ecliptic. 

57.  The  points,  where  the  ecliptic  intersects  the 
equator,  are  called  the  equinoctial  points  ;  or  the  equi- 
noxes. The  point  through  which  the  sun  ascends  from 
the  southern  to  the  northern  side  of  the  equator  is  called 
the  vernal  equinox,  and  the  other  equinox  is  called  the 
autumnal  equinox.  % 

The  points  90°  distant  from  the  ecliptic  are  called 
the  solstitial  points,  or  the  solstices.  [B.  p.  49.] 

58.  The  circumference  of  the  ecliptic  is  divided  into 
twelve  equal  parts,  called  signs,  beginning  with  the 
vernal  equinox,  and  proceeding  with  the  sun  from  west 
to  east. 

The  names  of  these  signs  are  Aries  (°f),  Taurus  (8), 
Gemini  (n),  Cancer  (zb),  Leo  (g\,,)  Virgo  (v%),  Libra  (£t), 


262  SPHERICAL    ASTRONOMY.  [CH.  V. 

Colures.  Tropics.  Latitude  of  a  star. 

Scorpio  (Trl),  Sagittarius  (f),  Capricornus  (V?),  Aquari* 
us  (£#),  Pisces  (X)-  The  vernal  equinox  is  therefore  the 
first  point,  or  beginning  of  Aries,  and  the  autumnal  equinox 
is  the  first  point  of  Libra;  the  first  six  signs  are  north  of  the 
equator,  and  the  last  six  south  of  the  equator.  The  northern 
solstice  is  the  first  point  of  Cancer,  and  the  southern  solstice 
the  first  point  of  Capricorn.    [B.  p.  49.] 

59.  Secondary  circles  drawn  perpendicular  to  the 
ecliptic  are  called  circles  of  latitude. 

The  circle  of  latitude  drawn  through  the  equinoxes 
is  called  the  equinoctial  colure. 

The  circle  of  latitude  drawn  through  the  solstices  is 
called  the  solstitial  colure.    [B.  p.  49.] 

Corollary.  The  solstitial  colure  is  also  a  secondary  to  the 
equator,  so  that  it  passes  through  the  poles  of  both  the  equator 
and  the  ecliptic. 

60.  Small  circles,  drawn  parallel  to  the  equator 
through  the  solstitial  points,  are  called  tropics. 

The  northern  tropic  is  called  the  tropic  of  Cancer  ; 
the  southern  tropic  the  tropic  of  Capricorn. 

Small  circles,  drawn  at  the  same  distance  from  the 
poles  which  the  tropics  are  from  the  equator,  are  called 
polar  circles. 

The  northern  polar  circle  is  called  the  arctic  circle, 
the  southern  the  antartic. 

61.  The  latitude  of  a  star  is  its  distance  from  the 
ecliptic  measured  upon  the   circle  of  latitude,  which 


$  64]  THE    ECLIPTIC.  263 

Longitude  of  a  star.  Nonagesimal  point. 

passes  through  the  star.  If  the  observer  is  supposed  to 
be  at  the  earth,  the  latitude  is  called  geocentric  latitude  ; 
but  if  he  is  at  the  sun,  it  is  heliocentric  latitude.  [B. 
p.  49.] 

62.  The  longitude  of  a  star  is  the  arc  of  the  ecliptic 
contained  between  the  circle  of  latitude  drawn  through 
the  star  and  the  vernal  equinox.     [B.  p.  50.] 

Corollary.  The   longitude  and  right  ascension  of  the  first 
point  of  Cancer  are  each  equal   to  6A,  and  those  of  the  first 
point  of  Capricorn  are  each  equal  to  18A. 

63.  The  nonagesimal  point  of  the  ecliptic  is  the 
highest  point  at  any  time. 

Corollary.  The  distance  of  the  nonagesimal  from  the  zenith 
is  therefore  equal  to  the  distance  of  the  zenith  from  the  eclip- 
tic, that  is,  to  the  celestial  latitude  of  the  zenith;  and  the 
longitude  of  the  nonagesimal  is  the  celestial  longitude  of  the 
zenith. 


64.  Problem.  To  find  the  latitude  and  longitude  of 
a  star,  when  its  right  ascension  and  declination  are 
known. 

Solution.  Let  P  (fig.  35.)  be  the  north  pole  of  the  equator, 
Z  the  north  pole  of  the  ecliptic,  and  B  the  star.  Then  EQW 
will  be  the  equator,  NESW  the  ecliptic,  and  NPZS  the 
solstitial  colure,  so  that  the  point  S  is  the  southern  solstice, 
and  N  the  northern  solstice.  Now  if  the  arc  PB  be  produced 
to  cut  the  equator  at  M,  and  ZB  to  cut  the  ecliptic  at  L ; 
the  angle  ZPB  is  measured   by  the  arc  QM,  that  is,  by  the 


264  SPHERICAL    ASTRONOMV.  [CH.  V. 

To  find  a  star's  latitude  and  longitude. 

difference  of  the  right  ascensions  of  Q  and  M,  or  by  the  dif- 
ference of  the  ^fc's  right  ascension  and  18*;  that  is, 

ZPB  —  18*  —  R.  A.  z=  24*  —  (G*  +  R.  A.)      (411) 
or  =  R.  A.  —  18*  =  (R.  A.  +  6*)  —  24* 

or  =  24*  +  R.  A.  —  18*  —  R.  A.  +  6*. 

In  the  same  way 

PZB  —  NL  —  Long.  —  90°  (412) 

or  =  360°  —  (Long.  90°) 

or  =  —  (Long.  —  90°), 

in  which  the  first  values  of  ZPB  and  PZD  correspond  to 
the  star's  being  east  of  the  solstitial  colure  ;  the  second  and 
third  values  to  the  star's  being  west  of  the  colure.  We  also 
have 

PB  —  90°  —  Dec.  (413) 

BZ  =  90°  —  Lat.  (414) 

+  PZ  ==  90°  —  ZQ=QS 

=  obliquity  of  ecliptic  z=  ±  E,     (415) 

in  which  the  declination  and  latitude  are  positive  when  north, 
and  neo-ative  when  south,  and  E  has  the  same  sign  with 
R.  A.  —  12*. 

The  present  problem   does   not,  then,  differ   from   that  of 
§  28,  and  if  we  put 

±i  =  PC-  90°,  (416) 

in  which  the  upper  sign  is  used,  when  R.  A.  —  12*  is  positive, 
and  otherwise  the  lower  sign,  we  have  by  (99,  105,  and  98) 

tang.  PC=n  =F  cotan.  A  =  cos.  (R.  A.  -\-  6*)  cotan.  Dec. 

b=  —  sin.  R.  A.  cotan.  Dec.      (417) 


§  65.]  THE    ECLIPTIC.  265 

To  find  the  latitude  and  longitude  of  a  star. 

in  which  the  signs  are  used  as  in  (416) ;  so  that  A  and  Dec. 
are  always  positive  or  negative  at  the  same  time.  Instead  of 
(417),  its  reciprocal  may  be  used,  which  is 

=p  tang.  A  —  —  cosec.  R.  A.  tang.  Dec.  (418) 

If,  then,  B  =  E  +  A,  (419) 

we  have 

AP  z=  =F  E  —  90°  =|=  A  =z  =p  B  —  90°  (420) 
or           z=  90°  ±  A  ±  -E  =  90°  ±  #, 

in  which  the  upper  or  lower  signs  are  used,  as  in  (415).  Hence 
cos.  PC :  cos.  AP  =  ^p  sin.  A  :  =p  sin.  B  —  sin.  A  :  sin.  J5 

z=  sin.  Dec.  :  sin.  Lat.  (421) 

so  that,  since  Dec.  and  A  are  both  positive  or  both  negative, 
B  and  Lat.  must  also  be  both  positive  or  both  negative.  Again, 

sin.  PC  :  sin.  PA  =  cos.  A  :  i  cos.  B  (422) 

=  dzcotang.  (R.  A  +  6A) :  ±  cotan.  (Long.  —90°) 
=r  zh  tang.  R.  A.  :  ±  tang.  Long. 

in  which  the  signs  may  be  neglected,  and  Long,  is  to  be  found 
in  the  same  quadrant  with  R.  A.,  unless  the  foot  P  of  the 
perpendicular  falls  within  the  triangle;  in  which  case  the  first 
value  of  AP  (420)  is  used,  so  that  B  is  obtuse.  In  this  case, 
the  longitude  is  in  the  adjacent  quadrant  on  the  same  side  of 
the  solsticial  colure  with  the  right  ascension.  These  results 
agree  with  the  Rule  in  [B.  p.  435.] 

65.  Corollary,    The  latitude  and   longitude  of  the  zenith, 
that  is,  the  zenith  distance  and   longitude  of  the  nonagesimal, 
might  be  found  by  the  same  method.     But  another  rule  can  be 
23 


266  SPHERICAL    ASTRONOMF.  [CH.   V. 

To  find  the  latitude  and  longitude  of  a  star. 

used,  which  is  of  peculiar  advantage,  where  these  quantities 
are  often  to  be  calculated  for  the  same  place.  We  have  by 
(310)  and  (311),  calling  B  the  zenith,  and  putting 

T  =  24*  —  ZPB  or  —  ZPB  (423  ) 

F=£(PZB  —  ZBP)  or  ==  180°—  £ (PZB—ZBP)  (424) 

G  —  i  (PZB  +  ZBP)  or  =  180°—  J  (PZB+  ZBP)  (425) 

tang.P^--  cosec.J(PJ5+PZ)sin.  J(PjB— PZ)cotan.JZT 

=  tang.  (24*  —  P)  (426) 

tang. 6?  =  —  sec.  £  (PP+PZ)cos.J(PP— PZ)cot.  J P  (427) 

90° -f  P+#=  PZB +  90°  or  =360°  —  PZB -{-90°  (428) 

=  Long,  or  =  360°  +  Long.  (429) 

in  which  the  first  member  of  (426)  is  used  when  PB  is 
greater  than  PZ,  and  the  third  when  PB  is  less  than  PZ, 
that  is,  within  the  north  polar  circle ;  and  the  second  members 
of  (423,  424,  425,  428)  correspond  to  the  position  of  the  ze- 
nith at  the  east  of  the  solstitial  colure,  but  the  third  members 
to  the  west  of  the  colure. 

Again,  by  (295), 

tang.  J  (90°  —  lat.)  z=r  tang.  J  alt.  nonagesimal 

=  cos.  G .  sec.  F  tang.  §  (PB  +  PZ),         (430) 

and  the  preceding  formulas  correspond  to  the  rule  in  [B. 
p.  402.] 

66.  Scholium.    The  rule  with  regard  to  the  value  of  G  ap- 
pears to  be  a  little  different,  but  the  difference  is  only  apparent ; 


$  68.]  THE    ECLIPTIC.  267 

To  find  the  altitude  of  the  nonagesimal. 

for  it  follows  from  (427),  that  G  and   12A  —  J  T  are,  at  the 
same  time,  both  acute  or  both  obtuse,  unless 

i  (PB  +  PZ)>  90°, 

or  PjB>180°  —  PZ,  (431) 

which  corresponds  to  the  south  polar  circle. 

67.  The  abridged  method  of  calculating  the  altitude  and 
longitude  of  the  nonagesimal  [B.  p.  403.],  only  consists  in  the 
previous  computation  of  the  values 

A  —  log.  [cos. (*-PB  —  PZ)  sec. i(PB  +  PZ)]    (432) 
C  z=  log.  tang,  i  (PB  +  PZ)  (433) 

B  =  log.  tang.  £  (PB  —  PZ)  —  C  (434) 

=z  log.  [tang.  £  (PB  —  PZ)  cotan.  J-  (PB  +  PZ)] 
==  log.  [co8ec.£(PB  +  PZ)Bin.£(PB—PZ)]—A, 
whence 

log.  [cosec.i(PB-{-PZ)sm.i(PB— PZ)]  —  B  +  A  (435) 

and  log.  tang.  G  =  A  +  log.  (-—  cotan.  |  T)  (436) 

log.  tang.  F=A  +  B  +  log.  (— cotan.£  T)       (437) 

=  log.  tang.  G  -\-  B 

log.  tang.  £  alt.  non.  =  log.  cos.  (2  -f-  log.  sec.  F+  C      (438) 

68.  The  rule  in  [B.  p.  436.]  for  finding  right  ascension  and 
declination,  when  the  longitude  and  latitude  are  given,  may 
be  obtained  by  a  process  precisely  similar  to  that  for  the  rule 
before  it. 


268  SPHERICAL    ASTRONOMY.  [CH.  V. 

To  find  the  latitude  and  longitude  of  a  star. 


69.    Examples. 

1.  Calculate  the  latitude  and  longitude  of  the  moon,  when 
its  right  ascension  is  4A  42OT  56%  and  its  declination  27°  21' 
58"  N.,  and  the  obliquity  of  the  ecliptic  23°  27'  45". 

Solution.  27°  21'  58"  N.  tang.  9.71400 

4A  42™  56s  tang.  0.45650         cosec.  0.02503 

A  =  28°  44'  12"  N.       sec.  0.05708  tang.  9.73903 

E  ==  23°  27'  45"  S. 


B  =    5°  16/27"N.       cos.  9.99816  tang.  8.96524 


long.  =  72°  53'  31"  tang.  0.51174  sin.  9.98034 


lat.=    5°    2'33"N.  tang.  8.94558 

2.  Calculate  the  values  of  A,  B,  and  C  for  the  obliquity 
23°  27'  40",  and  the  reduced  latitude  of  42°  12'  2"  N. 

Solution.  Polar  dist.  ==  47°  47'  58" 

47°  47'  58" 
23°  27'  40"  . 


£  sum  =z  35°  37'  49"     sec.  0.09002     tang.  9.85536 
diff.  =  12°  10'    9"     cos.  9.99013     tang.  9.33374 


A  =  0.08015,    B  =  9.47838 

3.  Calculate  the  altitude  and  longitude  of  the  nonagesimal, 
when  the  right  ascension  of  the  meridian  is  19*  50™,  the  lati- 
tude 42°  12'  2"  N.,  and  the  obliquity  23°  27'  40". 


<§>  69.]  THE   ECLIPTIC.  269 

To  find  the  latitude  and  longitude  of  a  star. 

Solution.     T  =  19*  50™  +  6*  —  24*  =  1*  50- 
£(l*50m)         cotan.  0.61137 
A  ss;  0.08015 


£=101°  30'    2"     tang.  0.69152        cos.  9.29968 
90°  B  =  9.4783S       C=  9.85536 


F=  124°    4'    5"     tang.  0.16990        sec.  0.25167 


long.  ==  315°  34'    7"  14°  18'  40"    tang.  9.40671 

alt.  =  28°  37'  20". 

4.  Calculate  the  latitude  and  longitude  of  the  moon,  when 
its  right  ascension  is  18*  27m  12s,  and  its  declination  27°  49 
38"  S.,  and  the  obliquity  of  the  ecliptic  23°  27'  45". 

Ans.     The  i)  's  long.  ==  276°    1'  46" 

its  lat.       =      4°30/26"S. 

5.  Calculate  the  values  of  A,  B,  and  C  for  Albany,  and 
the  obliquity  23°  27'  40". 

Ans.  A  =3  0.07967 
B  —  9.47573 
C  =  9.85333 

6.  Calculate  the  longitude  and  altitude  of  the  nonagesimal, 
when  the  obliquity  of  the  ecliptic  is  23°  27'  40",  the  latitude 
42°  12'  2"N.,  and  the  R.  A.  of  the  meridian  10*  10". 

Ans.     The  long.  =  138°  30'  23" 

alt.     =     61°  18'  49". 
23* 


270  SPHERICAL    ASTRONOMY.  [CH.     V. 


To  find  the  declination  of  a  star. 


7.  Calculate  the  moon's  right  ascension  and  declination, 
when  its  latitude  is  5°  0'  1"  N.,  its  longitude  64°  54'  1",  and 
the  obliquity  of  the  ecliptic  23°  27'  45". 

Ans.         Its  R.  A.  -=    4h  7m  46*. 

Its  Dec.    =  26°  3'    V  N. 

70.  Problem,    To  find  the  declination  of  a  star. 

Solution.  I.  Observe  its  meridian  altitude,  and  its  declina- 
tion is  at  once  found  by  one  of  the  equations  [345  -  347.] 

II.  If  the  star  does  not  set,  and  both  its  transits  are  observ- 
ed, we  have 

p  —  90°  —  Dec.  =  J  (A  j  — -  A').  (438) 

71.  Problem.  To  find  the  position  of  the  equinoctial 
points. 

Solution.  Since  the  right  ascension  of  all  stars  is  counted 
from  the  vernal  equinox,  and  since  the  two  equinoxes  are  12* 
apart,  the  present  problem  is  the  same  as  to  find  the  right 
ascension  of  some  one  of  the  stars,  which  may  afterwards 
serve  as  a  fixed  point  for  determining  the  right  ascension  of 
the  other  stars. 

Observe  the  declination  of  the  sun  for  several  successive 
noons  near  the  equinox,  until  two  noons  are  found  between 
which  its  declination  has  changed  its  sign ;  and  observe  also 
the  instant  of  the  sun's  transit  across  the  meridian  on  these 
days,  by  a  clock  whose  rate  of  going  is  known.  Then,  by 
supposing  the  sun's  motions  in  declination  and  right  ascension 
to  be  uniform  at  this  time,  which  they  nearly  are,  the  time 
of  the  equinox,  that  is,  of  the  sun's  being  in  the  equator,  is 
found  by  the  proportion 


§  72.]  THE    ECLIPTIC.  271 

To  find  the.  right  ascension  of  a  star. 

the  whole  change  of  declination  :  either  declination  = 
the  sideral  interval  between  the  transits  —  24/l  :  the 
sideral  interval  between  the  transit  of  the  equinox  and 
that  of  the  sun  at  this  declination;  (439) 

and  this  interval  is  the  difference  between  the  right  ascensions 
of  the  sun  at  this  declination  and  the  equinox.  If  the  passage 
of  a  star  had  been  observed  in  the  same  day,  the  right  ascen- 
sion of  the  star  would  have  been  the  interval  of  sideral  time 
of  its  passage  after  that  of  the  vernal  equinox.  t 

72.    Examples. 

1.    If  the    sun's  declination  is   found  at  one  transit  to  be 
7'  9".5  S.,  and  at  the  next  transit  to  be  16'  3I".l  N. ;  what  is 
the  sun's  right   ascension  at  the  second  transit,  if  the  sideral 
interval  of  the  transits  is  24/l  3™  38s.21. 
Solution. 

T  9".5  +  16'  31".  1  =  23'  40//.6  —  1420'.6  ar.co.  6.84753 
1G'  31  .1  ae  991".l  2.99612 

3m  38?.21  =  218s.21  2.33887 


©'s  R.  A.  ==  0h  2"  32s.2         152\2  2.18252 

2.  If  the  sun's  declination  is  found  at  one  transit  to  be 
18'  38".8S.,  and  at  the  next  transit  to  be  5'  3".2  N.  j  what  is 
the  sun's  right  ascension  at  the  second  transit,  if  the  sideral 
interval  of  the  transits  is  24*  3™  38s.4  ? 

Ans.     0h  0m  46*.5. 

3.  If  the  sun's  declination  is  found  at  one  transit  to  be 
5'  57".9  N.,  and  at  the  next  transit  to  be  17'  26//.3  S. ;  what  is 
the  sun's  right  ascension  at  the  second  transit,  if  the  sideral 
interval  of  the  transits  is  24A  3™  35s.71 1 

Ans.  12*2™40*.8. 


272  SPHERICAL    ASTRONOMY.  [CH.  V. 

To  find  the  obliquity  of  the  ecliptic. 

73.  Problem.    To  find  the  obliquity  of  the  ecliptic. 

Solution.  I.  Observe  the  right  ascension  and  declination 
of  the  sun,  when  he  is  nearly  at  his  greatest  declination  ;  that 
is,  when  his  right  ascension  is  nearly  6h  or  18\  If  he  were 
observed  at  exactly  his  greatest  declination,  the  observed 
declination  would  obviously  be  the  required  obliquity.  But  for 
any  other  time,  the  sun's  declination  and  right  ascension  are 
the  legs  of  a  right  triangle,  of  which  the  obliquity  of  the  eclip- 
tic is  the  angle  opposite  the  declination.     Hence 

tang.  ©'s  Dec.  =z  sin.  ©'s  R.  A.  tang,  obliq.         (440) 

Now  if  we  put 

h  =  the  diff.  of  ©'s  R.  A.  and  R.  A.  of  solstice, 

we  have 

7        tang,  ©'s  Dec.  #***, 

cos.  h  =  ^^4-_ —  (441 

tang.  Obliq.  v        ' 

and  by  (277)  and  (278), 

sin.  (obliq.  —  ©'s  dec.)  _    I — cos.  h 2 sin.2  %h 

sin.  (obliq.  -j-  ©'s  dec.)  ~~  1-f-cos.  h       2 cos.  2%h 

—  tang.2  J  £  (442) 

sin.  (obliq.  —  ©'s  dec.)  =  (obliq.  —  ©'s  dec.)  sin.  \" 

=  tang.2  £  /Vsin.  (obi.  +  ©'s  dec.)  (443) 

obi.  —  ©'s  dec.=  cosec.  I" tan.2  £ h sin.  (obl.+  ©'s  dec.)  (444) 

— -i/^cosec.l^tan.2  Is  sin.  (obi.  -f  ©'sdec.) 

and  the  second  member  of  (444)  may  be  regarded  as  a  cor- 
rection in  seconds  to  be  added  to  the  ©'s  dec.  to  obtain  the 
obliquity,  and  the  obliquity  in  the  second  member  need  only 
be  known  approximately. 


<§>  74.]  THE    ECLIPTIC.  273 


To  find  the  obliquity  of  the  ecliplic. 


74.    Examples. 

1.  The  right  ascensions  and  declinations  of  the  sun  on  sev- 
eral successive  days  were  as  follows  : 

June  19,  R.  A.  =  5*  50™ 53%  Dec.  —  23°26'  45".2N. 

20  5  55     3  23  27  27  .3 

21  5  59  12  23  27  44  .7 

22  6     3  21  23  27  37  .3 

23  6     7  31  23  27     4  .6 

To  find  the  obliquity  of  the  ecliptic. 

Solution.  Assume  for  the  obliquity  the  greatest  observed 
declination,  or  23°  27'  45",  and  the  corrections  of  all  the  ob- 
servations may  be  computed  in  the  same  way  as  that  of  the 
first,  which  is  thus  found, 

I  cosec.  1"  tang.2  Is  =  *§*  tang.  1"  6.43570 

h  —  9m  T  =  547s  2  log.  5.47598 

23°  26'  45"  +  23°  27'  45"  z=  4G°  54'  30"     sin.  9.86348 


cor.  dec.  —  59".59  1.77516 

23°  26'  45".2 


obliquity  —  23°  27'  44".8  =z  23°  27'  44".8 

In  the  same  way  the  2d  observation  gives  23  27  44  .9 

the  3d  observation  gives  23  27  45  .2 

the  4th  observation  gives  23  27  45  .3 

the  5th  observation  gives  23  27  45  .3 

sumzzr  117  18  45  .5 


The     mean  =  23°  27'  45".  1 


274  SPHERICAL    ASTRONOMY.  [CH.  V. 

To  find  the  obliquity  of  the  ecliptic.' 

2.  The  right  ascensions  and  declinations  of  the  sun  on  sev- 
eral successive  days,  were  as  follows : 

Dec.  20  ©'s  R.  A.  =  llh  51m  1 4s  23°  26'  48".4  S. 

21  17  55  40  23  27  30  .0 

22  18     0     7  23  27  44 .0 

23  18     4  33  23  27  29  .5 

24  IS     9    0  23  26  45  .5 

what  was  the  obliquity? 

Ans.     23°  27'  44//.7. 


§   76.]  PRECESSION    AND    NUTATION.  275 

Secular  and  periodical  motions. 


CHAPTER    VI. 


PRECESSION    AND    NUTATION. 


74.  The  ecliptic  is  not  a  fixed  but  a  moving  plane, 
and  its  observed  position  in  the  year  1750  has  been 
adopted  by  astronomers  as  a  fixed  plane,  to  which  its 
situation  at  any  other  time  is  referred. 

The  motion  of  the  ecliptic  is  shown  by  the  changes  in  the 
latitudes  of  the  stars. 

75.  Celestial  motions  are  generally  separated  into  two 
portions,  secular  and  periodical. 

Secular  motions  are  those  portions  of  the  celestial 
motions  which  either  remain  nearly  unchanged,  or  else 
are  subject  to  a  nearly  uniform  increase  or  diminution 
which  lasts  for  so  many  ages,  that  their  limits  and  times 
of  duration  have  not  yet  been  determined  with  any 
accuracy. 

Periodical  motions  are  those  whose  limits  are  small, 
and  periods  so  short,  that  they  have  been  determined 
with  considerable  accuracy. 

76.  The  true  position  of  a  heavenly  body,  or  of  a 
celestial  plane,  is  that  which  it  actually  has ;  its  mean 


276  SPHERICAL    ASTRONOMY".  [CH.  VI. 

Position  of  the  mean  ecliptic. 

position  is  that,  which  it  would  have  if  it  were  freed 
from  the  effects  of  its  periodical  motions. 

The  mean  position  is,  consequently,  subject  to  all  the  secu- 
lar changes. 

77.  The  mean  ecliptic  has,  from  the  time  of  the 
earliest  observations,  been  approaching  the  plane  of  the 
equator  at  a  little  less  than  the  half  of  a  second  each 
year,  thus  causing  a  diminution  of  the  obliquity  of  the 
ecliptic. 

Let  NAA1  (fig.  41.)  be  the  fixed  plane  of  1750,  and  NAX 
the  mean  ecliptic  for  the  number  of  years  t  after  1750.  Let 
A  be  the  vernal  equinox  of  1750,  and  AQ  the  equator.     Let 

n  z=  NA  and  n  =  the  angle  ANA  1  ; 

then,  upon  the  authority  of  Bessel,  the  point  of  intersection  N 
of  the  ecliptic,  which  is  called  the  node  of  the  ecliptic,  with 
the  fixed  plane,  has  a  retrograde  motion,  by  which  it  ap- 
proaches A  at  the  annual  rate  of  5".  18,  and  if  this  point  could 
have  existed  in  1750,  its  longitude  would  have  been  171°  36' 
10",  so  that 

§     ii—  171°  36'  10"  —  5".  18  t.  (445) 

Moreover,  the  angle  which  the  mean  ecliptic  makes  with  the 
fixed  plane  increases  at  the  annual  rate  of  0".48892,  but  this 
rate  of  increase  is  itself  decreasing  at  such  a  rate,  that  at  the 
time  t  this  angle  is 

rr  —  0",4S892  t  —  0".00000307l9  t*  (446) 

78.  Problem.  To  find  the  change  of  the  mean  lati- 
tude of  a  star,  which  arises  from  the  motion  of  the 
ecliptic. 


$  78.]  PRECESSION    AND    NUTATION.  277 

Change  of  mean  latitude. 

Solution.     Let 

L  =  the  #'slat.  in  1750 
3  L  =  its  change  of  lat. 
J.  z=  its  long,  in  1750—171°  36'  10"  +  5".  18*  (447) 

=  its  long,  referred  to  the  node  of  the  ecliptic 
<M  z=  its  change  of  long,  from  the  node ; 

then,  if  Z  (fig.  42)  is  the  pole  of  the  fixed  plane,  P  that  of 
the  ecliptic,  and  B  the  star ;  we  have 

PZ  =tt,  ZB  =  90°  —  L,  PB  =  90°  —  L  —  dL 

PZB  =  90°  +  j,  P  =  90°  —  J  —  9 A 

Draw  ZC  perpendicular  to  PB,  and  we  have,  since  PZ,  PC, 
and  CZ  are  very  small, 

PC  —  PZ  cos.  P  =  it  sin.  (^  +  9  A) 
or  ■=.  TV  sin.  J- 

cos.  PZ  :  cos.  PC  =  cos.  PZ  :  cos.  BC 
or  P^  =  PC 

PC—  PB  —  BZ=z—dL=iTvSm.j 
3  L  =  —  re  sin.  A  (448) 

S3  _  (0".4S892  *  —  0//.0000030719  *2)  sin.  >. 
Again,  the  triangle  ZPP  gives,  by  (295), 
sin. \(PZB+P) :  cos.|(PZP- P)=:tan.jU :  tan.£(PP+P,Z) 
But 

£(PZB  +  P)  =  90°  — %U,  i(PZB—P)  =  J+ltj, 
24 


278  .  SPHERICAL   ASTRONOMY.  [CH.  VI. 

Mean  celestial  equator. 

whence  9  A  —  n  cos.  4  tang.  L  (449) 

sa  (0".48892  t  —  0".0000030719  t2)  cos.  a  tang.  L. 

79.  The  mean  celestial  equator  has  a  motion  by  which 
its  node  upon  the  fixed  plane  moves  from  the  node  of 
the  ecliptic  at  the  annual  rate  of  about  50",  while  its 
inclination  to  the  fixed  plane  has  a  very  small  increase 
proportioned  to  the  square  of  the  time  from  1750. 

Thus,  if  AQ  (fig.  41.)  is  the  equator  of  1750,  and  A1  Q' 
that  for  the  time  f,  so  that  A  is  the  vernal  equinox  of  1750, 
and  AP A  x  that  for  the  time  t. 

Let  yj  =  AA't  to  —  NA'Qi, 

then  A'  moves  from  A  at  the  annual  rate  of  50 '.340499,  and 
this  rate  is  diminishing  so  that  at  the  time 

y  =  50//.340499  t  —  O'.OOO  121 7945  t2,  (450) 

and  the  value  of  w  in  the  year  1750  was 

*  —  23°  28'  18", 

and  is  increasing  at  a  rate  proportioned  to  the  square  of  the 
time,  so  that 

w  =  co/  -J-  0". 00000984233  t2.  (451) 

80.  Problem.   To  find  the  change  of  the  mean  ob- 
liquity of  the  ecliptic  and  that  of  longitude. 

Solution.     Let  (fig.  41.) 

JV^Q'z^,     NAA1  =  tl  +  Ji; 


§  80.]  PRECESSION    AND    NUTATION.  279 

Change  of  mean  obliquity  and  longitude. 

then,  by  (310)  and  (311), 

sin.  [#+£(v  + Vi)]  _  tang.fr(q)-f  cnj 


(452) 
(453) 


sin.  J  (V — V^)  tang.  J  n 

cos,  [g  +  ^^  +  ^a)]  __  tang.  1(g)!  -co) 
cos.  £  (v —  Vj)  tang.  £  7i 

Now  in  calculating  the  parts  of  y1  —  %p  and  Wj  — to,  which 
are  proportional  to  the  time,  we  may,  since  y  and  xp1  differ  but 
little  as  well  as  w  and  Ml9  and  since  n  is  small,  put 

2r-fi(v  +  ^i)=  ?*  sin-  5-(^  —  V'1)  =  ^(V  —  vjsin.  1" 
tang.  |  n  —  £  7T  tang.  1"  —  £  tt  sin.  1"  =  ^  (0".48892)  t1  sin.  1" 

£(«  +  »!)=«',  tang.JK—  w)  =  i(wi—  w)sin.l' 
cos.  £  (v  —  v^)  p  1, 
which,  subtituted  in  (452)  and  (453),  give 

yj  —  yj1z=z  (T. 48892  t  sin.  tt  cotan. »'  (454) 

Wl  —  co  —  0  ".48892  *  cos.  n,  (455) 

which  are  thus  computed, 

0".48S92  9.68924  9.68924 

171°  36' 10"     cos.  9.99532,        sin.     9.16446 


—  0".48368  9.68456n 

23°  28'  18"     cotan.  0.36229 


0".  164431 

9.21599 

that  is,  »,  —  a,  =  —  0".48368  t 

(456) 

y  —  y1—     0".  164431* 

(457) 

280  SPHERICAL    ASTRONOMY.  [CH.  VI, 

Change  of  mean  obliquity  and  longitude. 

or      mi=z  23°  28'  18"  —  0  '.48368  I  (458) 

fl  =~  50".340499  t  —  0".l  64431  t  —  50M76068  t.  (459) 

But,  in  computing  the  parts  of  Wj  —  w  and  y  —  y%9  which 
depend  upon  t2}  we  need  only  retain  the  part  depending  upon 
t2  in  the  value  of  tang.  J  n,  and  neglect  these  pajrts  in  the 
other  terms  of  (452)  and  (453),  we  thus  have 

sin.  [tz+  -J-  (v  +  vi )]  =  sin,  (n  +  45".08 1)        (460) 
z=  sin.  n-\-  45". 08  £  sin.  i".  cos.  n 
cos.[^4-^(^  +  ^1)]z=cos.(n)—45//.08/sin.  l"sin.  n  (461) 
'tan.^=ffi  sin.  I"=£sin.  l//(0//.48892  If  —  0".00000307 1 9  *2  (462) 

cotan.  £  (co  -f  ij)  —  cotan.  (o>  —  0  .24184  *,        (463) 

_  l-j-0'.24184*sin.  Ftang  «' 
Z  tang.  W  —  0 "34184  t  sin.  1" 

=  cotan.  J  +  0' .24184  t  sin.  1"  ( I  +  cotan.2  o') 

=  cotan.  a/  -f-  0//.24L84  *  sin.  \"  cosec.2  ^ 

cos.  J  (v —  ¥t)  =  l|  sin.  ^  (v  —  Y^)  ==  £  (^  —  Vjsin.  I" 
sin.  £  (Wj  —  w)  =  J  (Wj  —  w)  sin.  1" 
which,  substituted  in  (452)  and  453),  give 
v,— 1//1=0//. 164431  *  +  0//.48892*2  sin.  1"45".08 cos. n cotan. «>' 
-f  0 '.48892  *2  sin.  1"  X  0//.24184sin.  n cosec.2  w      (464) 
—  (K0000030719  t2  sin.  77  cotan.  to' 
to  t  —  cd  —  _  0  ".48368  *  —  (K48892*2  sin.  1"45".08  sin.  n 
—  0/ .0000030719  t2  cos.  77, 


<§>  80.]  PRECESSION    AND    NUTATION.  281 

Change  of  mean  obliquity  and  longitude. 

which  are  thus  computed, 

0".48892  9.68924 

1"  sin.  4.68557 

45".08  1.65398 

171°  36'  10"  sin.  9.16446  cos.  9.99532n 


— -0".  000015605  5.19325 

+  0^.000003039  0".  0000030719      4.48741 


— 0".000012566  4.48273R 

0".0000030719         4.48741 

171°  36' 10"       sin.  9.16446  cos.  9.99532n      sin.  9.16446 

23°  28'  18"  cotan.  0.36229  0.36229  cosec*  0.79958 


— 0".000001033 

4.01416 
45".08 

0".48892 

0".24184 

sin. 

1 

'  4.68557 
1.65398 
9.68924 

4.68557 
9.68924 

— 0".000243445 

6.38640,, 

9.38353 

0".000000528 

3.72238 

—0.000243950 

so  that      y  —  xp  1  =  0".  164431 1  —  0".000243950  t* 
«1— toz=—  0".48368  t  —  0".000012566  W 
Vjt  =z  50".176068  *  —  0".0001217945 12  +  0".000243950 1* 

—  50".176068  t  -f  0".000122156  t*  (465) 

w  x  zz:  23°  28'  18"  —  0".48368  t  —  0".000002724  *2         (466) 
24* 


282  SPHERICAL    ASTRONOMY.  [CH.  VI. 

Precession  of  the  equinoxes. 

or  more  accurately,  from  BesseFs  Fundament  a  Astronomic, 
V ',  =s=  50".176068*  +  O'.OOO  1221483  t2  (467) 

w,  =±  23°  28'  18"  —  0".48368  t  —  0". 00000272295  t      (468) 

These   values  were    afterward    changed    by   Bessel   in  his 
TabulcB  RegiomontancB  to 

V  =  50".37572  *  —  0'  .0001217945  t2  (469) 

y,  z=  50".21129  t  +  0'.0001221483  tf  .  (470) 

to,  =z  23°  28' 18"  —  0".48368f  —  0//.00000272295  ^  (471) 

Bat  these  formulas  were  obtained  from  the  physical  theory, 
and  are,  as  Bessel  says,  subject  to  errors,  on  account  of  the 
uncertainty  with  regard  to  some  of  the  data  ;  so  that  we  shall 
adopt  Poisson's  formulas,  because  they  agree  in  the  variation 
of  the  obliquity  almost  exactly  with  BesseFs  observations,  and 
shall  change  the  value  of  »'  to  that  determined  by  Bessel  from 
observations;  our  formulas  are,  then, 

w'  =  23°28'17".65  (472) 

y  =  50".37572  t  —  0.//00010905  t2  -  (473) 

V I  =  50//.22300  t  +  0."0001  1637  t2  (474) 

w  ~  23°  28'  17".65  -f  0".0000800i  t2  (475) 

»t  =  23°  28'  17".65  —  0".45692  t  —  0".000002242 12  (476) 

If,  now,  the  value  of  yx  is  added  to  that  of  d  4  (449),  the 
resulting  value  is  the  total  change  of  a  star's  mean  longitude. 

81.  The  backward   motion  yt  of  the   equinoxes  is 
called  the  precession  of  the  equinoxes. 


*82.] 


PRECESSION    AND    NUTATION. 


283 


Change  of  mean  equator. 


82.  Problem.    To  find  the  intersection  of  the  mean 
equator  with  the  equator  of  1750  and  its  inclination  to  it. 

Solution.  Produce  A  Q  and  AQ'  (fig.  41.)  till  they  meet  at 
T,  and  let 

AT—  *,  AT—  <*>', 

and  the  triangle  ATA1  gives,  by  (291,  295,  and  310), 

cos.  J  (to7  —  w) :  cos. J  (a/  -{-  w)=ztang. J  v  :  tang. £(<*>' — <£>)  (477) 

sin.  £  (ex)'  —  w)  :  sin. J  (a/  -|-  w)z=tang.J  v  :  tang. J( *'-)-*)  (478) 

sin.  J  (#'-(-#):  sin.J  (</>'—*):=:  cot  an.  J  J7 :  cot.  $  (o'-fco)  (479) 

so  that  t2  may  be  neglected  in   all   the  terms  but  y,  and  we 
have 

1  :  cos.  »'  ±=  £  V»  sin.  1"  :  |  (*'  —  <?>)  sin.  1"       (480) 

0  :  sin.  J  =  £  y  sin.  1"  :  tang.  £  (<£'+<*>)  (481) 

1  :  |  (**  —  «£)  sin.  1"  =  tang.  j'i  J.  T  sin.  1".     (482) 
Hence  £  (*'  +  *)  =  90°  (483) 

i  (*'  —  tf>)  =  £  *  cos.  co7  (4S4) 

y  =  (<*>/  __  i)  tang,  co7,  (485) 

which  are  thus  computed, 

a,'  cos.  9.96249         cos.  9.96249 


25"  18786 

1.40120 

23".  103 

1.36369 

0".000054525 

5.73660 

o".oooo50oi3 

5.69909 

co> 

tang. 

9.63771 

9.63771 

10//.032 

1.00140 

0".000021717 

5.33680 

284  SPHERICAL    ASTRONOMY.  [cH.  VI. 

Change  of  mean  right  ascension  and  declination. 

so  that 

&  =  90°  —  23".  103 1  +  0'.000050013  t*         (486) 

T  —  20".  0640 1  —  0".000043434 t2.  (487) 

83.  Problem.  To  Jind  the  variation  of  a  star's  mean 
right  ascension  and  declination. 

I.  The  variation,  which  arises  from  the  change  of  the  equa- 
tor's inclination,  may  be  found  precisely  in  the  same  way  in 
which  the  variations  of  latitude  and  longitude  were  found  in 
§  78,  for  a  similar  change  in  the  position  of  the  ecliptic ;  so 
that  formulas  (448)  and  (449)  give,  by  substituting  for  -^,  L 
and  7i, 

A  =  #'s  R.  A.  —  90°  +  23".103  t  =:  R  —  90 

L  =  #'s  Dec.  =  D,     tv~  T 
3D  z=  —  T  cos.  R  (488) 

8R~  T  sin.  R  tang.  D  ;  (489) 

or  instead  of  counting  the  value  of  T  and  t  from  1750,  they 
may  be  reduced  to  the  beginning  of  each  year,  and  the  square 
of  t  may  then  be  neglected. 

II.  The  variation  in  right  ascension  is  to  be  increased  by 
the  change  in  the  position  of  the  equinox,  arising  from  its 
precession,  which  is  thus  found.  Had  the  ecliptic  remained 
stationary,  the  equinox  would  have  removed  from  A  to  A\  so 
that  if  AP  is  perpendicular  to  the  equator,  we  should  have 
for  the  increase  of  right  ascension  by  (475)  and  (484), 

AP  =  AA!  cos.  AAP  =  y  cos.  «  (490) 

=  (*'  —  <*>) 

=  46//.206  t  —  G^.000100026  t2. 


<§>  84.]  PRECESSION    AND    NUTATION.  285 

Change  of  mean  right  ascension. 


But  the  equinox  advances  upon  the  equator  from  the  motion 
of  the  ecliptic  by  the  arc  A'Alt  which  is  thus  found.  We 
have,  by  (291), 

cos.^-(co1  —  w):cos.]J(a)1  +  co)  =  tang.  £  4'^  :tang.  £  (y—  yj 

But  COS.  z(o)1 w)  zzz   1 

cos.  £  (co1  -f  w)  =  COS.  (a;'  —  0".22846  * ) 

—  cos.  a/  +  0'.22846 1  sin.  1"  sin.  J 

sec.  2-  (wj  +  w)  =  sec.  «'  —  0 '.22846 1  sin.  1"  sin.  w'  sec.2  w' 
tang.  £  il'J ,  =  i  yi'4  j  sin.  I" 

tang.£(v/  —  Vi|=  i(v  —  Vi)  sin-  j" 

==  J  sin.  1"  (0".15272  *  —  0".00022542*2) 
whence      A,A1  =  0".  15272 1  sec.  w 

—  0' .00022542  *2  sec.  */ 

—  0' .22846 t2  0;  15272 t2  sin.l"  tang.  w>  secV 
which  is  thus  computed, 

0  ".15272  9.18390  9.18390 

sec.  0.03751  0.03751  0.03751 


0U665  9.22141 


0".00022542     6.35299 


0,/.00024575  6.39050 

0".22846  9.35881 

tang.  9.63771 

1"  sin.  4.68557 


2.90350 


286  SPHERICAL    ASTRONOMY.  [CH.  VI. 

Nutation. 

so  that 

A' A  t  —  0".1665  t  —  0".00024575  t2,  (491) 

and,  by  (489)  and  (490), 
d  R  —  46".0395 1  +  0".00016593 12  +  Tain.  R  tang,  D.  (492) 

84.  By  the  motions  of  precession  and  of  diminution 
of  the  obliquity,  the  mean  pole  of  the  equator  is  carried 
round  the  pole  of  the  ecliptic,  gradually  approaching 
it ;  but  the  true  pole  of  the  equator  has  a  motion  round 
the  mean  pole,  which  is  called  nutation.  This  motion 
is  in  an  oval,  at  the  centre  of  which  is  the  mean  pole, 
and  is  such  that  the  position  of  the  mean  equinox  dif- 
fers from  that  of  the  true  equinox  by  the  longitude 

d  \ong.=i  sin.  £1  +  ^  sin.  2  Sl+i2  sin.  2j>  +  i8  sin.2©  (493) 

where 

£l  =  the  mean  longitude  of  that  point  of  intersec- 
tion of  the  moon's  orbit  with  the  ecliptic, 
through  which  the  moon  ascends  from 
the  south  to  the  north  side  of  the  ecliptic, 
and  which  is  called  the  moon's  ascend- 
ing node, 

J)  =  the  moon's  true  longitude, 

©  =.  the  sun's  true  longitude. 

The  values  of  i,  ilf  i2,  i3  are  given  differently  by  different 
astronomers,  and  those  which  are,  at  present,  adopted  in  the 
Nautical  Almanac  are 

i  —  —  17'.2985,     ix  =  0".2082  (494) 

i2=  —  0".2074,      »8  =  —  1 '.2550. 


§  86.]  PRECESSION    AND    NUTATION.  287 

Nutation. 

This  nutation  of  the  pole  causes  also  the  true  obliquity  of  the 
ecliptic  to  change  from  the  mean  obliquity  by  the  quantity 

^1=kcos.^l  +  k1cos.2£i-\-k2cos.2j>  +  kscos.2Q  (495) 

in  which  the  values  of  k  &c,  at  present  adopted  in  the  Nauti- 
cal Almanac,  are 

k   &  9".2500,     k1=z  —  0".0903  (496) 

k2  =  0".0900,     k3  =  0".5447. 

85.  Corollary.  The  effect  of  nutation  upon  the  right 
ascensions  and  declinations  of  the  stars  may  be  com- 
puted by  §  83,  and  the  formulas  which  are  obtained 
agree  with  those  given  in  the  Nautical  Almanac,  and 
which  depend  upon  the  terms,  called  C  and  D  in  the 
formulas  for  Reduction  of  the  Almanac  ;  these  terms 
contain  also  the  changes  arising  from  the  mean  motion 
of  the  equinoxes,  and  the  formulas  are  so  reduced  that 
t  is  counted  from  the  beginning  of  each  year. 


86.    Examples. 

1.  Find  the  mean  obliquity  of  the  ecliptic  for  the  year  1840, 
and  reduce  the  formulas  for  finding  the  variations  of  right  as- 
cension and  declination  to  the  beginning  of  that  year. 

Solution.    In  (476)     let      t  =  1840  —  1750  =  90, 

and  it  gives 

m%  =  23°  28'  17 ".65  —  41".12  —  0'.02  =:  23°  27'  36' .51. 

In  (487,  488,  and  492)  let  t  hn  90  +  *',  and  neglect  the 
terms  depending  upon  t'2t  so  that 


288  SPHERICAL    ASTRONOMY.  [CH.  VI. 

Change  in  right  ascension  and  declination. 

T  =  30'  5  ".76  —  0'.35  +  20".0640 11  —  0".0078 11 

z=30' 5".41  +  20".0562t", 

and  the   mean  variations,  counted  from  the   beginning  of  the 
year,  are 

*  D  —  20 ".0562 1  cos.  R 

$  R  =  46//.0693  t1  +  20  ".0562  t1  sin.  R  tang.  D. 

Finally,  the  variations  arising  from  nutation  are  thus  found. 
The  change  in  the  obliquity  of  the  ecliptic  gives  at  once,  from 
(448)  and  (449),  by  referring  the  positions  to  the  mean  eclip- 
tic instead  of  to  that  of  1750, 

*  D  =  —  aWl  sin.  R 

d'  R  =  —  <5  Wj  cos.  R  tang.  D, 

and  the  change  in  the  position  of  the  equinox  gives  by  (485, 
488,  489,  and  490), 

2T=  —  a  A  sin.  Wl 
d'D=  $  A  sin.  m1  cos.  J? 
d'  Rz=z  3  A  cos.  w  x  -{-  $  A  sin.  w1  sin.  12  tang.  D. 

Hence,  if  we  take 

46//.0693  C  =  46 ".0693  t"  +  <M  cos.  », 

c  ==  46".0693  +  20".0562  sin.  12  tang.  1> 
c'=  20".0562  cos.JR 
d  —  cos.  R  tang.  X) 
e£  =  —  sin  R 

we  have  °  =  ^  +  46^693  ^  =  ^  +  2^^562  ^ 

—  te  —  0.3448  sin.  &  +  0.00415  sin.  2  & 
—  0.00413  sin.  2  J>  —  0.02502  sin. 2  O, 


<§>  86.]  PRECESSION    AND    NUTATION.  289 

Nutation  in  right  ascension  and  declination. 


and  the  entire  changes  of  declination  and  right  ascension  are 

^  D  —  Cc'  —  da.d' 

V  R  —.  Cc  —du.d, 

which  agree  with  the  formulas  in  the  Nautical  Almanac,  ex- 
cept in  the  coefficients  of  V ,  which  are  46".0206  and  20".0426 
instead  of  46//.0693  and  20".0562. 

If,  again,  we  take 

/  ==  46".0693  C, 
g  cos.  G  =  20//.0562  C,    g  sin.  G  =  —  d  », 
the  above  formulas  become 

d'  D=  gcos.  Gcos.R — gsin.  6rsin.  R  =:  g  cos.  (G + R) 
<*' R  —  f-{-gsm. R cos.  Gtang.  D-\-g sin.  Gcos.  R tang.  D 
=f+g  sin.  (R  +  G)  tang.  D, 
as  in  the  Nautical  Almanac. 

2.  Find  the  annual  variations  in  the  right  ascension  and 
declination  of  «  Hydrae  for  the  year  1840,  and  its  true  place 
for  mean  midnight  at  Greenwich,  Jan.  1,  1840;  its  mean  right 
ascension  for  Jan.  1,  1839,  being  9*  19w40*.620,  and  its  decli- 
nation —  7°  57'  49 '.50,  and  using  the  numbers  of  the  Nautical 
Almanac. 


25 


290 


SPHERICAL    ASTRONOMY. 


[CH.  VI. 


Nutation  in  right  ascension  and  declination. 

Solution. 

20".0426  1.30195 

R=z9h  19CT  40*.620     cos.  9.8S374* 

(jD=-  15".335  1.18569" 

D  =  —  7°  57  49".50 

* R  =  46".0206  —  1' .8051 
=  44  ".2155  =  2'.948 


1.30195 

sin.  9.80872 

tang.  9.14584" 


0.25651* 


Hence  its  mean  place  for  Jan.  1,  1840,  is 
R  —  9h  19-  43s.568 
D  —-T  58'  4".83. 
To  calculate  the  effects  of  nutation,  we  have 

a  =  339°  40,    J)  =  242°  30',     ©  =  281°  15' 

—0.3448    sin.    &=     0.1205,      9".25     cos.    &  =     8".673 

0.00415  sin.  2  &  =—0.0027,  —  0".0903  cos.  2  &  =— 0".068 

—0.00413  sin.  2  J>  =— 0.0034,      0".0900  cos.  2  J>  =— 0 ".032 

—0.02502  sin.  2  ©  =     0.0096,      0".5447  cos.  2  ©  =— 0".504 

C  =  f  +     0.1240,  <5Wl-     8".049 

Pc;=?  c'  f  -f  20".0426  X  0.1240  cos.  R 

=  c't'  —  15".335  X  0.1240  z=  c <  t>  —  1".901 
—  *  ».<*'  =  8' .049  sin.  JE  -z  5".  181 
Cc  =  c  t>  -f  0.1240  X  2*.948  =  ct'-\-  0*.365 
—  I  w d  =  —  8' .049  cos.  JR  tang. Dm — 0".861  =  —  0*.058, 


§  86.]  PRECESSION    AND    NUTATION.  291 

Nutation  in  right  ascension  and  declination. 


whence  the  variations  arising  from  nutation  are 

d<  D  =  3' .28,     d '  R  =  0*.30, 
and  the  true  places  are 

D  =  —  7°  58'  1'.55,     R  =  9*  19"  43s.87. 

3.  Find  the  mean  obliquity  of  the  ecliptic  for  the  year  1950, 
and  reduce  the  formulas  for  finding  the  variations  of  mean  right 
ascension  and  declination  to  the  beginning  of  that  year. 

Ans.     Wl  =  23°  26'  36".18. 
*'  D  =  19".8903 11  cos.  R 
*'  Rz=l  46".  1059  t1  +  197/.8903  t  sin.  R  tang.  2>. 

4.  Find  the  annual  variations  in  the  right  ascension  and 
declination  of  /J  Ursse  Minoris  for  the  year  1839,  and  its  true 
place  for  mean  midnight  at  Greenwich,  Aug.  9,  1839;  its 
mean  right  ascension  for  Jan.  1,  1839,  being  14A  5lm  14s.943, 
its  declination  74°  487  48//.89  N.,  the  longitude  of  the  moon's 
ascending  node  for  Aug.  9,  1839,  being  347°  177,  that  of  the 
moon  144°  2;,  and  that  of  the  sun  136°  307,  and  using  the 
constants  of  the  Nautical  Almanac,  which  give  for  Aug.  9, 
1839, 

f—  32//.33,    g  —  16 '.70,     G  —  327°  30'. 

Ans.    Var.  in  R.  A.  —~ 0'.277 ;  var.  in  Dec.  =  U'.ll ; 
and  for  Aug.  9,  1839, 

R  a  14A  51CT  16*.36 
D  =  74°  48'  32//.46. 


292  SPHERICAL    ASTRONOMY.  [CH.  VI. 

Tables  XL  and  XLIII. 


5.  Calculate  the  values  of  ft  g,  and  G  for  April  1,  1839, 
mean  midnight  at  Greenwich,  when  g\,  ==  354°  10',  ©  =  11° 
34',  and  J)  is  neglected. 

Ans.    f—  12'. 53,    g  —  1I"&*,     G  z=  299°  34. 

In  Table  XL  of  the  Navigator,  the  decimal  is  neglected, 
and  20  used  instead  of  20.0562.  Table  XLIII  is  calculated 
from  the  formulas  of  Bessel,  which  differ  a  little  from  those  of 
Bailly  used  in  the  Nautical  Almanac.  The  construction  of 
these  two  tables  is  sufficiently  simple  from  the  calculations 
already  given. 


§  89.]  time.  293 

Sideral  and  solar  day. 


CHAPTER    VII. 

TIME. 

87.  The  intervals  between  the  successive  returns  of 
the  mean  place  of  a  star  to  the  meridian  are  precisely 
equal,  and  the  mean  daily  motion  of  the  star  is  perfectly 
uniform  ;  so  that  sideral  time  is  adapted  to  all  the  wants 
of  astronomy.  The  instant,  which  has  been  adopted 
as  the  commencement  of  the  sideral  day,  is  the  upper 
transit  of  the  vernal  equinox. 

The  length  of  the  sideral  day,  which  is  thus  adopted,  differs 
therefore  from  the  true  sideral  or  star  day  by  the  daily  change 
in  the  right  ascension  of  the  vernal  equinox.  But  this  change 
is  annually  about  507/  or  3S.3,  so  that  the  daily  change  is  less 
than  O'.Ol,  and  is  altogether  insensible. 

88.  Corollary.  The  difference  between  the  sideral 
time  of  different  places  is  exactly  equal  to  the  differ- 
ence of  the  longitude  of  the  places. 

89.  The  interval  between  two  successive  upper  tran- 
sits of  the  sun  over  the  meridian  is  called  a  solar  day  ; 
and  the  hour  angle  of  the  sun  is  called  solar  time. 
This  is  the  measure  of  time  best  fitted  to  the  common 
purposes  of  life. 

25* 


294  SPHERICAL    ASTRONOMY.  [CH.   VII. 

Perigee.  Apogee. 

The  intervals  between  the  successive  returns  of  the  sun  to 
the  meridian  are  not  exactly  equal,  but  depend  upon  the  vari- 
able motion  of  the  sun  in  right  ascension,  and  can  only  be 
determined  by  an  accurate  knowledge  of  this  motion. 

90.  The  want  of  uniformity  in  the  sun's  motion  in 
right  ascension  arises  from  two  different  causes. 

I.  The  sun  does  not  move  in  the  equator  but  in  the 
ecliptic. 

II.  The  sun's  motion  in  the  ecliptic  is  not  uniform. 
The  variable  motion  of  the  sun  along  the  ecliptic,  and  its 

deviations  from  the  plane  of  the  mean  ecliptic,  cannot  be  dis- 
tinctly represented,  without  reference  to  the  variations  of  its 
distance  from  the  earth,  and  to  the  nature  of  the  curve  which 
it  describes.  This  portion  of  the  subject,  therefore,  which 
involves  the  determination  of  the  sun's  exact  daily  position, 
that  is,  the  calculation  of  its  ephcmeris,  must  be  reserved  for 
the  Physical  Astronomy.  It  is  sufficient,  for  our  present 
purpose,  to  know  that  the  sun  moves  with  the  greatest  velocity 
when  it  is  nearest  the  earth,  that  is,  in  lis  perigee;  and  that  it 
moves  most  slowly  when  it  is  farthest  from  the  earth,  that  is,  in 
its  apogee. 

91.  The  sun  arrives  at  its  perigee  about  8  days  after 
the  winter  solstice,  and  at  its  apogee  about  8  days  after 
the  summer  solstice.  The  mean  longitude  of  the 
perigee  at  the  beginning  of  the  year  1800  was  279° 
30'  5",  and  it  is  advancing  towards  the  eastward  at  the 
annual  rate  of  about  11".  8,  so  that,  by  adding  the  pre- 
cession of  the  equinoxes,  the  annual  increase  of  its 
longitude  is  about  62". 


$  95.]  time.  295 

Mean  and  apparent  time;  equation  of  time. 

92.  To  avoid  the  irregularity  of  time  arising  from  the 
want  of  uniformity  of  the  sun's  motion,  a  fictitious  sun, 
called  a  mean  sun,  is  supposed  to  move  uniformly  in 
the  ecliptic  at  such  a  rate,  as  to  return  to  the  perigee  at 
the  same  time  with  the  true  sun.  A  second  mean  sun 
is  also  supposed  to  move  in  the  equator  at  the  same 
rate  with  the  first  mean  sun,  and  to  return  to  each 
equinox  at  the  same  time  with  the  first  mean  sun. 

We  shall  denote  the  first  mean  sun  by  Q  1}  and  the  second 
mean  sun  by  ©2. 

93.  Corollary.  The  right  ascension  of  the  second 
mean  sun  is  equal  to  the  longitude  of  the  first  mean 
sun. 

94.  The  time  which  is  denoted  by  the  second  mean 
sun  is  perfectly  uniform  in  its  increase,  and  is  called 
?nean  time;  while  that  which  is  denoted  by  the  true 
sun  is  called  true  or  apparent  time;  the  difference  be- 
tween mean  and  true  time  is  called  the  equation  of 
time. 

95.  The  time  which  it  takes  the  sun  to  complete  a 
revolution  about  the  earth  is  called  a  year. 

The  time  which  it  takes  the  mean  sun  to  return  to 
the  same  longitude  is  the  common  or  tropical  year. 

The  time  which  it  takes  it  to  return  to  the  same  star 
is  the  sideral  year ;  and  the  time  which  it  takes  it  to 
return  to  the  perigee  is  the  anomalistic  year. 


296  SPHERICAL   ASTRONOMY.  [cH.  VII. 

Year.  Leap  year. 


The  length  of  the  mean  tropical  year  is 

F  =  365d  5h  48"  4V.808,  (497) 

so  that  the  daily  mean  motion  of  the  sjn  is  found  by  the  pro- 
portion 

Y  :  ld  =  360°  :  daily  motion  =  59'  8  ".3302.        (498) 

96.  The  fraction  of  a  day  is  necessarily  neglected  in  the 
length  of  the  year  in  common  life,  and  the  common  year  is 
taken  equal  to  365d.  By  this  approximation,  the  error  in  four 
years  amounts  to 

23*  15-  1P.232  =  ld  —  Um  48*. 768,  (499) 

or  nearly  a  day,  and  an  additional  day  is  consequently  added 
to  the  fourth  year,  which  is  called  the  leap  year.  At  the  end 
of  a  century  the  remaining  error  amounts  to  nearly  —  0d.75, 
which  is  noticed  by  the  neglect  of  three  leap  years  in  four 
centuries.  For  practical  convenience,  those  years  are  taken 
as  leap  years  which  are  exactly  divisible  by  4,  and  the  centu- 
rial  years  would  thus  be  leap  years,  but  only  those  are  re- 
tained as  leap  years  which  are  d. visible  by  400. 

97.  When  the  mean  sun  has  returned  to  the  same  mean 
longitude,  it  has  not  returned  to  the  same  star,  because  the 
equinox  from  which  the  longitude  is  counted  has  retrograded 
by  50" .223,  so  that  the  mean  sun  has  50;  .223  farther  to  go, 
and  the  time  of  describing  this  arc  is  the  fourth  term  of  the 
proportion 

59'  8'.3302  :  ld  =  50//.223  :  20m  22a.786,  (500) 

so  that  the  length  of  the  sideral  year  is 

Yt  =  Y  +  20*  22s.786  =  365d  6h  9m  10'.594.       (501) 


<§>  98.]  time.  297 

Tables  LI  and  LII.         Reduction  of  solar  to  sideral  time. 

98.  The  length  of  the  mean  solar  day  is  also  different  from 
that  of  the  sideral  day,  because  when  the  ©2,  in  its  diurnal 
motion,  returns  to  the  meridian,  it  is  59/  87/.3302  advanced 
in  right  ascension  ;  so  that  360°  59'  87/.3302  pass  the  meridian 
in  a  solar  day,  instead  of  360°,  which  pass  in  a  sideral  day. 
Hence  the  excess  of  the  solar  day  above  the  sideral  day,  ex- 
pressed in  solar  time,  is  the  fourth  term  of  the  proportion 

360°  59'  8".3302  :  59'  S'.3302  —  ld  :  0d.0027305 

or     Zh  55-.9094 ;  (502) 

that  is,  1  sid.  day  =  0.9972695  sol.  day, 

or  24A  sid.  time  a=  23*  56m  4s.O906  of  solar  time  ;  (503) 

which  agrees  with  (335)  and  the  table  for  changing  sideral  to 
solar  time  in  the  Nautical  Almanac,  and  with  table  LII  of  the 
Navigator. 

In  the  same  way  this  excess  expressed  in  sideral  time  is 
the  fourth  term  of  the  proportion 

360°  :  59'  8V3302  =  ld  :  0d.002738  or  3m  56s5554 ; 

that  is,  1  sol.  day  =  1.002738  sid.  day,  (504) 

or  24*  sol.  time  ==  24*  Sm  56s.5554  sid.  time  j         (505) 

which  agrees  with  the  table  for  changing  solar  to  sideral  time 
in  the  Nautical  Almanac,  and  with  table  LI  of  the  Navigator. 
The  remainder  of  tables  LI  and  LII,  as  well  as  the  corre- 
sponding ones  given  in  the  Nautical  Almanac,  are  calculated 
by  simple  proportions  from  the  numbers  which  are  given  for 
24\ 

The   sideral  day  begins  with  the  transit  of  the  true  vernal 


298  SPHERICAL    ASTRONOMY.  [CH.  VII. 

Sun's  longitude  on  January  1. 

equinox.     At  the  time  of  the  transit  of  ©2,   then,  that  is,  at 
mean  noon,  we  have 

the  sid.  time  :=  R.  A.  of  ©2  from  the  equinox 
zs  R.  A.  of  ©2  from  mean  equinox 

-}-  Nutation  of  equinox  in  R.  A. 
=2  sun's  mean  long -[-Nutation  in  R.  A.  (506) 

99.  The  sun's  mean  long,  for  Jan.  1,  1800,  at  Paris,  was 
found  by  Bessel  to  be  279°  54'  11  ".36.  Its  longitude  for 
Jan.  1,  of  any  other  year  t,  may  thus  be  found.  Let  f  be 
the  remainder  after  the  division  of  t  by  4,  the  number  of  days, 
then,  by  which  Jan.  1  of  the  year  t  is  removed  from  Jan.  1, 
1800,  is 

365^  (t  —f)  +  365/  z=  t.  365^  -  \  f 

ss  Y.t  +  t.llm\2°.l92  —  if         (507) 
ss  Y.  t  +  t .  0d.00778  —  if. 

But  in  Yt  days  the  sun's  longitude  increases  exactly  £.360°, 
which  is  to  be  neglected ;  and  its  increase  in  longitude  is 

59'  8".3302  (*+0.00778— lf)—t.21"M— f.  14' 47 ".083,  (508) 

or  more  accurately  from  Bes  el,  the  mean  longitude  E,  for  the 
the  first  of  January  of  the  year  1800  +  '  at  Paris,  is 

E  =  279°  54'  1  ".36  +  1 27".605844  + 1*.  0".0001221805 
— /,  14'  47".083.  (509) 

The  mean  longitude  is  found  for  the  first  of  January,  for 
any  other  meridian  by  the  following  proportion,  derived  from 
the  interval  of  time  between  the  ©2>s  passage  over  this  me- 
ridian and  that  of  Paris. 

24* :  long,  from  Par.  ss  59/8//.3302  :  change  in  value  of  E.  (510) 


$  100.]  time.  299 

Sideral  time  converted  to  solar  time. 

The  sun's  mean  longitude  for  any  mean  noon  n  of  the  year 
after  that  of  the  first  of  Jan.  is 

E  +  n.59'  8'.3302.  (511) 

Hence  the  sideral  time  of  the  mean  noon  n  is 
•pi 
&=  -  +  ».3W  56^.555348  +  Nutation  in  R.  A.  (512) 

so  that  the  solar   time  of  the  transit  of  the  equinox  from  the 
preceding  noon  is 

24*  —  S  (converted  into  solar  time).  (513) 

100.    Examples. 

1.  Find  the   sideral  interval   which  corresponds  to  10*  of 
solar  time. 

Ans.     10   lm38s.5647. 

2.'  Find  the  solar  interval  which  corresponds  to  10*  of  si- 
deral time. 

Ans.     9*  58"  2K7044. 

3.  Find  the  sideral  interval   which  corresponds  to  10w  of 
solar  time. 

Ans.     10m  P.6428. 

4.  Find  the  solar  interval  which  corresponds  to  10*  of  si- 
deral time. 

Ans.     9m  5S'.3617. 

5.  Find  the   sideral    interval  which  corresponds  to  10*  of 
solar  time. 

Ans.     1C.0274. 


300  SPHERICAL    ASTRONOMY.  [CH.  VII. 

'     Time  by  observation  of  equal  altitudes. 

6.  Find  the  solar  interval  which  corresponds  to  10*  of  si- 

deral  time. 

Ans.     9*.9727. 


7.  Find  the  sideral  interval  which  corresponds  to  05.85  of 

solar  time. 

Ans.     C.85233. 

8.  Find   the  solar   interval  which   corresponds  to  0S.85  of 

sideral  time. 

Ans.     0S.84768. 

9.  Find  the  sun's  mean  longitude  at  Greenwich  for  the 
mean  noon  of  April  4,  1839,  the  sideral  time  at  this  noon,  and 
the  solar  time  of  the  transit  of  the  vernal  equinox  from  the 
preceding  noon;  the  meridian  of  Greenwich  is  9m  21s.5  wes, 
of  that  of  Paris. 

Ans.     The  sun's  mean  longitude  =  12°  T  3".02. 

The  sideral  time  of  mean  noon  =  48m  3P.27. 

Time  of  tran.  of  ver.  equi.  ==  April  3d,  23*  1  lm  39*.6S. 

101.  Problem.     To  find  the  time  by  observation. 
Solution.     First  Method.     By  equal  altitudes. 

I.  If  the  star  does  not  change  its  declination.  Observe  the 
times  when  the  star  is  at  equal  altitudes  before  and  after  pass- 
ing the  meridian ;  the  arithmetical  mean  between  these  two 
times  is  the  time  of  the  star's  passing  the  meridian,  which 
compared  with  the  known  time  of  this  passage,  gives  the  error 
of  the  clock  at  this  time,  and  the  correction  of  this  error  gives 
the  time  of  each  observation. 


$    101.]  TIME.  301 

Equal  altitudes  of  sun. 

IT.  When  the  declination  of  the  star  is  changing,  the  time 
of  the  star's  arriving  at  the  observed  altitude  A  is  affected ; 
thus  if 

L  sr  the  latitude, 

D  z=  the  declination  at  the  meridian, 

d  D  =i  the  increase  of  declination  from  the  meridian, 

h  —  the  hour  angle,  supposing  no  change  in  the  decli- 
nation, 

dh  —  the  increase  of  the  hour  angle  in  time, 

we  have,  by  (380), 

sin.  A  —  sin.  L  sin.  D  -f-  cos.  L  cos.  D  cos.  h         (514) 

=  sin.  L  s'm.(D-\- 3 D)  -\-  cos.L  cos.(D-{-d D)cos  (h-\-3h) 

=.  s'm.L  sin.D-{-dD  sin. I"  sin.L  cos.Z>-|-  cos.Z  cos.ZJcos.A 

—  3  D  sin.  1"  cos.Zsin.  Dcos.h — 15<j  h  sin.l"  cos.Z  cos.D  sin.  h, 

whence 

0  zz:  <J  D  sin.  L  cos.  D  —  $  D  cos  L  sin*  D  cos*  /* 

—  15  3  h  cos.  L  cos  D  sin.  h 

qJi=z  T^dD  tang.  L  cosec.  h  —  T\<$D  tang.  Z>  cotan.  A 


15  cotan.  Zi  sin.  h        15  cotan.  i>  tang.  K 


(515) 


and  since  the  two  observations  are  at  nearly  the  same  distance 
from  the  meridian,  the  value  of  S7i  is  the  same  for  both  of 
them  ;  so  that  their  mean  is  augmented  by  <J  h>  and  d  h  is 
consequently  to  be  subtracted  from  the  mean  of  the  observed 

26 


302  SPHERICAL    ASTRONOMY.  [CH.  VII. 

Equal  altitudes  of  sun. 

times,  in  order  to  obtain  the  true  time  of  the  star's  passing  the 
meridian. 

In  calculating  the  value  of  3  h,  its  two.  terms  may  be  calcu- 
lated separately.  Now  if  & D  is  the  daily  variation  of  the 
star's  declination,  we  have 

hi'D       2h#D  /e%aK 

'D  =  -*4T  -5X24*  (516) 

and  in  using  proportional  logarithms,  the  proportional  logarithm 
of  the  hours  and  minutes  of  2  h,  which  is  the  elapsed  time, 
may  be  taken  as  if  they  were  minutes  and  seconds,  provided 
the  same  is  done  with  the  2-ih  in  the  denominator.  Finally, 
the  value  of  <5  h  is  reduced  from  minutes  and  seconds  to  sec- 
onds and  thirds  by  multiplying  by  60,  so  that  if  M  is  taken 
for  the  denominator  of  either  of  the  parts  of  (515),  this  part  P 
is  calculated  by  the  formula 

Prop.  log.  P  = — Prop.  log.  g*2^*15  +  log.M+Prop.log.2  h 

+  Prop.  log.  6'  D,  (517) 

which  agrees  with  [B.  p.  219.],  for 

—Prop.  log.  *QA    -  =  — Prop.  log.  12-  =  —  1.1761 

=  8.8239.    (518) 

III.  If  the  altitude  at  the  two  observations  had  differed 
slightly,  the  mean  time  would  require  to  be  corrected  ;  for  this 
purpose,  let 

3  A  =  the  excess  of  the  second  altitude  above  the  first, 

3  h  z=  the  increase  of  the  hour  angle, 


§  101.]  TIME.  303 

Altitudes  nearly  equal.  Single  altitude. 

and  we  easily  deduce  from  (514) 

cos.  A  3  A  =z  —  15  cos.  L  cos.  D  sin.  h  dht         (519) 

xi.  ,  7  cos.  ^4  *  A  ,^™v 

so  that  s  h  =  —  — = pr^ — r.  (520) 

15  cos.  Lt  cos.  i>  sin.  Ii 

The  time  of  the  second  observation  being  thus  increased  by 
dh,  that  of  the  mean  is  increased  by  J  dh,  which  is,  therefore, 
the  correction  to  be  subtracted  from  this  mean. 

The  corrections  (515)  and  (520)  must  be  both  of  them  ap- 
plied when  the  star  is  changing  its  declination,  and  at  the 
same  time  the  observed  altitudes  are  slightly  different. 

Second  Method.     By  a  single  altitude.    [B.  p.  208-218.] 

When  a  single  altitude  is  observed,  there  are  known  in  the 
triangle  PZB  (fig.  35.),  the  three  sides,  to  find  the  hour  angle 
ZPB,  which  is  thus  found  by  (277), 

*  s  =  J  (z  +  90°  —  L  +  p)  (521) 

cos.J^v(-^^(^-V  (522) 

2  ^  \sin.(90°  —  JL)sin.p/'  v        ' 

which  corresponds  to  [B.  p.  210.] 

The  hour  angle  may  also  be  found  by  (282),  thus  if  we  put 
s<  =  i(A  +  L+p),  (523) 

we  have 
5  =  J  (180°  —  A  —  Z,+jp)=z90°— sf+pz=90o  —  A— L+s< 

s—p  =  90°  —  s'>  s  —  (90°  —  L)=zs'  —  A, 

whence 

•      -.t  ./'cos.  sf  sin.  (s*  —  A)\  /co<\ 

sin.  i  h  =  \/|  ^A -  I,  (524) 

2  ^  \       cos.  L  sin.  p       ] 

which  corresponds  to  [B.  p.  209.] 


304  SPHERICAL    ASTRONOMY.  [cH.  VII. 


Distance  from  terrestrial  object. 


Third  Method.     By  the  distance  from  a  fixed  terrestrial 
object. 

If  the  position  of  the  terrestrial  object  has  been  before  de- 
termined, its  hour  angle  and  polar  distance  may  be  considered 
as  known. 

Hence,  if  T  (fig.  40.)  is  the  position  of  the  terrestrial  object 
projected  upon  the  celestial  sphere,  P  the  pole,  and  S  the 
star.     Let  the  distance  TS  be  observed,  and  let 

PT=P,     PS  —  Pt     TS=d, 

TPZ  —  H,     TPS  =  h't     SPT—  h, 

s=zi(P+p  +  d),  (525) 


we  have 
or 


.iV=V(^L^^SL=i)\  (526) 

■  \  sin.  P  sin.  p  J  v       ' 

17/  ,/sin.s  sin.  (s — d)\  ,•*-* 

s.  i  h'  —  V  (  — : =-3 1  I ,  (527) 

*  \     sin.  P  sin.  y      /  v        ' 

h=z  H+h'. 

If  the  polar  distance  and  hour  angle  of  the  terrestrial  object 
is  not  known,  but  only  its  altitude  and  azimuth,  the  polar  dis- 
tance and  hour  angle  can  be  easily  found  by  solving  the  tri- 
angle PZT. 


Fcurth  Method.     By  a  meridian  transit.   [B.  p.  221.] 

If  the  passage  of  a  star  is  observed  over  the  different  wires 
of  a  transit  instrument,  the  mean  of  the  observed  times  is  the 
time  of  the   meridian  transit,  which   should   agree   with  the 


<§>  101.]  TIME.  305 

Meridian  transit.  Vertical  transit. 


known  time  of  this  transit.     This  method  surpasses  all  others 
in  accuracy  and  brevity. 

Fifth  Method.     By  a  disappearance  behind  a  terrestrial 
object. 

If  the  instant  of  a  star's  disappearance  behind  a  vertical 
tower  has  been  observed  repeatedly  with  great  care,  the  ob- 
served time  of  this  disappearance  may  afterwards  be  used  for 
correcting  the  chronometer.  For  this  purpose,  the  position  of 
the  observer  must  always  be  precisely  the  same.  Any  change 
in  the  right  ascension  of  the  star  does  not  affect  the  star's 
hour  angle,  that  is,  the  elapsed  time  from  the  meridian  transit ; 
this  change,  consequently,  affects  the  observed  time  exactly  as 
if  the  observation  were  that  of  a  meridian  transit. 

A  small  change  in  the  declination  of  the  star  affects  the 
hour  angle,  and  therefore  the  time  of  observation.  Thus,  if 
P  (fig.  44.)  is  the  pole,  Z  the  zenith,  ZSS1  the  vertical 
plane  of  the  terrestrial  object ;  then  if  the  polar  distance  PS 
is  diminished  by 

RS  —  3  D, 

the  hour  angle  ZPS  is  diminished  by  the  angle 

SS'P  —  $h. 

But  the  S'R  is  nearly  perpendicular  to  SP,  and  the  sides 
of  SS'R  are  so  small,  that  their  curvature  may  be  neglected, 
whence 

RS'  =  <*  D  tang.  S  =  15  cos.  D.  *  A, 

so  that  dhz=LT\dD  tang.  S  sec.  D.  (528) 

26* 


306  SPHERICAL   ASTRONOMY.  [cH.'viI. 

To  find  the  time. 


102.    Examples. 

1.  On  July  25,  1823,  in  latitude  54°  20'  N.,  the  sun  was  at 
equal  altitudes,  the  observed  interval  was  6h  lm  36* ;  find  the 
correction  for  the  mean  of  the  observed  t  mes.  The  sun's 
declination  is  19°  48',  and  his  daily  increase  of  declination 
12'  44 ". 

8.8239 


tang.  0.0030 
1.4759 
1.1503 


Solution. 

8.8239 

54°  20' 

cotan. 

9.8559, 

6*  lm  36' 

sin. 

9.8510 

6m  1* 

P.  L. 

1.4759 

12'  44" 

P.L. 

1.1503 

12*55 

1.1570 

— 2*.28 

—  2*.28     1.8968 


10*.3  =  the  required  correction. 

2.  On  September  1,  1824,  in  latitude  46°  50'  N.,  the  inter- 
val between  the  observations,  when  the  sun  was  at  equal  alti- 
tudes, was  7*  46'"  35*;  the  sun's  declination  was  8°  14' N., 
and  his  daily  increase  of  declination  — 21'  49";  what  is  the 
correction  for  the  mean  of  the  observations  ? 

Ans.     16*5. 

3.  On  March  5,  1825,  in  latitude  38°  34'  N.,  the  interval 
between  the  observations,  when  the  sun  was  at  equal  altitudes, 
was  8*  29™  28';  the  sun's  declination  was  6°  2' S.,  and  his 
daily  increase  of  declination  was  23' 9";  what  is  the  correction 
for  the  mean  of  the  observations? 

Ans.     15'.4. 


$  102.]  time.  307 

To  find  the  time. 


4.  On  March  27,  1794,  in  latitude  51°  32'  N.,  the  interval 
between  the  observations,  when  the  sun  was  at  equal  altitudes, 
was  7h  29*  55s;  the  sun's  declination  was  2°  47'  N.,  and  his 
daily  increase  of  declination  23'  26";  what  is  the  correction 
for  the  mean  of  the  observations  1 

Arts.     —  21'.7. 

5.  In  latitude  20°  26'  N.,  the  altitude  of  Aldebaran,  before 
arriving  at  the  meridian,  was  found  to  be  45°  20',  and,  after 
passing  the  meridian,  to  b?  45°  10';  the  interval  between  the 
observations  was  7h  16m  35%  and  the  declination  of  Aldebaran 
was  16°  10' N.  j  what  is  the  correction  tor  the  mem  of  the 
observations  ? 

Arts.     19s. 

6.  In  latitude  3G°  39'  S.,  the  sun's  correct  central  altitude 
was  found  to  be  10°  40',  when  his  declination  was  9°  27'  N. ; 
what  was  the  hour  angle  ?  , 

Ans.     7h  23w  51*. 

7.  In  latitude  13°  17'  N.,  the  sun's  correct  central  altitude 
was  found  to  be  36°  37%  when  his  declination  was  22°  10'  S.  ; 
what  was  the  hour  angle  ? 

Ans.     9*  17m  88. 

8.  In  latitude  50°  56'  17"  N.,  the  zenith  distance  of  a  ter- 
restrial object  was  found  to  be  90°  24'  28",  and  its  azimuth 
35°  47'  4"  from  the  south ;  what  were  its  polar  distance  and 
hour  angle  ? 

Ans.     Its  polar  distance  =s  121°  6'  43" 
Its  hour  angle       =  2*  52w  16s. 


308  SPHERICAL    ASTRONOMY.  [CH.  VII. 


To  find  the  time. 


9.  From  the  preceding  terrestrial  object  three  distances  of 
the  sun  were  found  to  be  78°  9'  26",  77°  39'  26",  and  77° 
29'  26",  when  his  declination  was  14°  7'  13"  S. ;  what  were 
the  sun's  hour  angles,  if  he  was  on  the  opposite  side  of  the 
meridian  from  the  terrestrial  object? 

Ans.    %h  45™  49s,  2h  43m  27%  and  2*  42w  39'. 


§   103.]  LONGITUDE.  309 

By  measurement,  signals,  chronometer. 


CHAPTER    VIII. 

LONGITUDE. 

103.  Problem.     To  find  the  longitude  of  a  place. 

First  Method.     By  terrestrial  measurement. 

If  the  longitude  of  a  place  is  known,  that  of  another  place, 
which  is  near  it,  can  be  found  by  measuring  the  bearing  and 
distance  ;  whence  the  difference  of  longitude  may  be  calcu- 
lated by  the  rules  already  given  in  Navigation. 

Second  Method.     By  signals. 

The  stars,  by  their  diurnal  motion,  pass  round  the  earth 
once  in  24  sideral  hours ;  hence  they  arrive  at  each  meridian 
by  a  difference  of  sideral  time  equal  to  the  difference  of  longi- 
tude. In  the  same  way,  the  sun  passes  round  the  earth  once 
in  24  solar  hours  ;  so  that  it  arrives  at  each  meridian  by  a 
difference  of  solar  time  equal  to  the  difference  of  longitude. 
The  difference  of  longitude  of  two  places  is,  consequently, 
equal  to  their  difference  of  time.  Now  if  any  signal,  as  the 
bursting  of  a  rocket,  is  observed  at  two  places  ;  the  instant  of 
this  event,  as  noticed  by  the  clocks  of  the  two  places,  gives 
their  difference  of  time. 

Third  Method.     By  a  chronometer. 

The  difference  of  time  of  two  places  can,  obviously,  be 
determined    by  carrying  a  chronometer,   whose  rate   is  well 


310  SPHERICAL    ASTRONOMY.  [CH.  VIII. 

By  eclipses,  moon's  transit. 

ascertained,  from  one  place  to  the  other  ;  and  if  the  chronom- 
eter did  not  change  its  rate  during  the  passage,  this  method 
would  be  perfectly  accurate. 

Fourth  Method.     By  an  eclipse  of  one  of  Jupiter's  satel- 
lites. [B.  p,  252.] 

The  signal  of  the  second  method  cannot  be  used,  when  the 
places  are  more  than  20  or  30  miles  apart;  and,  when  the 
distance  is  very  great,  a  celestial  signal  must  be  used,  such  as 
the  immersion  or  emersion  of  one  of  Jupiter's  satellites.  For 
this  purpose,  the  instant  when  any  such  event  would  happen  to 
an  observer  at  Greenwich  is  inserted  in  the  Nautical  Almanac ; 
and  the  observer  at  any  other  place  has  only  to  compare  the 
time  of  his  observation  with  that  of  the  Almanac  to  obtain  his 
longitude  from  Greenwich. 

Fifth  Method.     By  an  eclipse  of  the  moon.  [B.  p.  253.] 

The  beginning  or  ending  of  an  eclipse  of  the  moon  may 
also  be  substituted  for  the  signal  of  the  second  method  to  de- 
termine the  difference  of  time. 

Sixth  Method.  By  a  meridian  transit  of  the  moon.  [B.  p.  431.] 

The  motion  of  the  moon  is  so  rapid,  that  the  instant  of  its 
arrival  at  a  given  place  in  the  heavens  may  be  used  for  the 
signal.  Of  the  elements  of  its  position  its  right  ascension  is 
changing  most  rapidly,  and  this  element  is  easily  determined 
at  the  instant  of  its  passage  over  the  meridian  by  the  differ- 
ence of  time  between  its  passage  and  that  of  a  known  star. 
The  instant  of  Greenwich  time,  when  the  moon's  right  ascen- 
sion is  equal  to  the  observed  right  ascension,  might  be  deter- 


§  103.]  LONGITUDE.  311 

By  a  moon's  transit. 

jnined  from  the  right  ascension,  which  is  given  in  the  Nautical 
Almanac  for  every  hour.  But  this  computation  involves  the 
observation  of  the  solar  time,  whereas  the  observed  interval 
gives  at  once  the  sideral  time  of  the  observation. 

The  calculation  is  then  more  simple,  by  means  of  the  table 
of  Moon -Culminating  stars  given  in  the  Nautical  Almanac,  in 
which  the  right  ascensions  of  the  suitable  stars  and  of  the 
moon's  bright  limb  are  given  at  the  instant  of  their  upper 
transits  over  the  meridian  of  Greenwich,  and  also  the  right 
ascension  of  the  moon's  bright  limb  at  the  instant  of  its  lower 
transit.  Hence  the  difference  between  the  right  ascensions  of 
the  moon's  limb,  at  two  successive  transits,  is  the  change  of 
its  right  ascension  in  passing  from  the  meridian  of  Greenwich 
to  that  which  is  12*  from  Greenwich ;  so  that  if  the  motion  in 
right  ascension  were  perfectly  uniform,  the  right  ascension, 
which  corresponded  to  a  given  meridian,  or  the  meridian, 
which  corresponded  to  a  given  right  ascension,  might  be  found 
by  the  following  simple  proportion, 

12*  :  long,  of  place  =  diff.  of  right  ascensions  for  12*  : 

diff.  of  right  ascensions  for  long,  of  place,  (529) 

in  which  the  longitude  of  the  place  may  be  counted  from  the 
meridian  12*  from  that  of  Greenwich,  provided  the  change  of 
right  ascension  for  an  upper  transit  is  computed  from  the  pre- 
ceding right  ascension,  which  is  that  of  a  lower  transit  at 
Greenwich,  that  is,  if  the  place  is  in  east  longitude. 

Let  then         T  =r:  long.,  if  west, 

or  =12*  —  long,  (if  the  long,  is  east) ; 

and  let  A  z=  diff.  of  right  ascension  for  the  Greenwich 

transits,  which  immediately  precede  and 
follow  the  required  or  observed  transit, 


312  SPHERICAL    ASTRONOMY.  [CH.  VIII. 


By  a  moon's  transit. 


and  let  d  A  =  change  of  right  ascension  from  the  pre- 

ceding  Greenwich    transit    to   the  ob- 
served transit, 

and  we  have,  by  (529), 

12*  :  T—  A  :tA,  (530) 

A  T  12*  a  A 

whence  6  A  — -,    and   T  =  — j—,  (531) 

1/Z  A. 

and  if  T  is  reduced  to  seconds,  we  have 

iA  =  mm  <532> 

log.  «  A  =  log.  A  +  log.  T  +  (ar.  co.)  log.  43200 

—  ]og.  A  +  log.  T  +  5.36452  (533) 

J,       43200  M  §JL-' 

and  T  =s (534) 

log.  r  =  4.63548  -f  ar.  co.  log.  4  +  log.  S  A,     (535) 

and  formulas  (533)  and  (535)  agree  with  the  parts  of  the  rules 
in  the  Naviga  or,  which  depend  upon  A,  and  are  independent 
of  the  want  of  uniformity  in  the  moon's  motion. 

The  corrections  which  arise  from  the  change  of  the  moon's 
motion  may  be  calculated,  on  the  supposition  that  this  motion 
is  uniformly  increasing  or  decreasing,  so  that  the  mean  motion 
for  any  interval  is  equal  to  the  motion  which  it  has  at  the  mid- 
dle instant  of  that  interval.     If  we  put,  then, 

B  zz  the  increase  of  motion  in  12*,  (536) 

A  is  not  the  mean  daily  motion  for  the  interval  of  longitude  T 
and  the  instant  J  T  after  the  meridian  transit  at  Greenwich, 
but  for  the  interval  12*  and  the  instant  6*  after  this  transit. 
The  mean  daily  motion  for  the  instant  £  T  is,  therefore, 


§  103.]  LONGITUDE.  313 

By  moon's  transit.  Table  XLV. 

{6>-j  T)B 


(538) 


so  that  the  correction  for  A  is 

(6A--  jT)B  _         (21600  8  —  j  T)B 
12*  "~  ""43200  ■ 

and  the  correction  of  3  A  in  (532)  is 

r(2i600--jr)  r(4320o-r) 

'*- (43200 r        B~ 2  (43200)2~ ■B'(539) 

and  the  value  of  d  B  is  easily  calculated  and  put  into  tables, 
like  Table  XLV  of  the  Navigator. 

In  correcting  the  value  of  T  (534),  the  correction  of  <S  A  is 
to  be  computed  from  Table  XLV  by  means  of  the  approxi* 
mate  value  of  T,  and  the  correction  of  T  is  then  found  by  the 
formula  to  be 

,r=i™i!  (540) 

It  only  remains,  to  show  how  to  find  the  value  of  B  from 
the  Nautical  Almanac.  Now  if  A'  denotes  the  motion  in  right 
ascension  for  the  12A  interval  of  longitude,  which  precedes 
that  to  which  A  corresponds ;  and  if  A"  denotes  the  motion 
in  right  ascension  for  the  12 h  interval  of  longitude  which  fol- 
lows that  of  A  ;  we  have 

2  B  =  A' '  —  A' 

B  =  i  (A"  —  A%  (541) 

and   the  calculation  agrees  entirely  with  that  given  in  the 
Navigator. 

When  the  longitude  is  small,  or  nearly  12*,  the  correction 
for  the  variation  of  motion  may  be  neglected,  provided,  instead 

27 


314  SPHERICAL    ASTRONOMY.  [CH.  VIII. 

By  a  lunar  distance. 

of  A,  the  motion  is  used  which  corresponds  to  the  time  of  the 
nearest  Greenwich  transit.  Now,  in  the  Nautical  Almanac, 
this  motion  is  given  for  an  hour's  interval,  of  which  the  mid- 
dle instant  is  that  of  the  transit,  so  that  if  H  =  this  hourly 
motion,  the  motion  for  the  time  T  may  be  found  by  the  for- 
mula 

lh  :  TzzzH.dA, 

whence 

SAXlh  _  2600-  X  '  A 
T  a        _        _  F (542) 

log.  T=z  3.55630  +  log.  9  A  +  (ar.  co.)  log.  H,     (543) 

which  agrees  with  [B.  p.  432.] 

The  formula  (543)  may  be  rendered  more  correct,  if  the 
value  of  H  is  taken  for  the  instant  J  T  of  longitude  ;  and 
the  value  can  be  computed  precisely  in  the  same  way  in 
which  the  right  ascension  was  computed  for  the  time  Tt  by 
noticing  the  want  of  uniformity  in  its  increase  ;  and  the  for- 
mula thus  corrected  is  accurate  for  small  differences  of  longi- 
tude. 

Seventh  Method.     By  a  lunar  distance. 

The  distance  of  the  moon  from  the  sun  or  a  star  may  be 
used  as  the  signal ;  but  the  true  places  of  these  bodies  differ 
from  their  apparent  places,  as  will  be  shown  in  succeeding 
chapters,  so  that  the  observed  distance  requires  to  be  corrected  ; 
and  the  correction  cannot  be  found  without  knowing  the  alti- 
tudes of  the  bodies.  It  is  sufficient,  for  the  present  purpose, 
to  know  that  the  difference  between  the  true  and  apparent 
places  is  only  a  difference  of  altitude,  and  not  one  of  azimuth, 
and  that  the  apparent  place  of  the  sun  or  a  star  is  higher  than 
its  true  place,  while  that  of  the  moon  is  lower.     The  true  dis- 


$  103.]  LONGITUDE.  315 


By  a  lunar  distance. 


tance  may,  then,  be  calculated  from  the  observed  distance  by 
one  of  the  following  methods. 

I.  Let  Z  (fig.  45.)  be  the  zenith,  S  the  apparent  place  of 
the  sun  or  star,  and  S'  the  true  place,  M  the  apparent  place 
of  the  moon,  M'  the  true  place  ;  let 

a  as  the  star's  apparent  alt.  as  90°  — ■  ZS 
a'  as  its  true  alt.  as  90°  —  ZS1 
b  as  the.  moon's  app.  alt.  as  90°  —  ZM 
b'  —  its  true  alt.  =  90°  —  ZM1 
E  sa  the  app.  dist.  as  SM 
E'  as  the  true  dist.  —  S1  M' 

Z  as  the  angle  Z  . 

a  6  as  ATM7  as  b'  —  6 

Then  the  triangles  ZSM  and  Zff'Jf'  give,  by  (273), 

2(cos.  i  **=  cos'jE+cos-(a+6)^cos^/+COS-(a/+6/)  (544) 
^      '  ^     '  cos.  a  cos.  6  cos.  a1  cos.  &' 

cos.  a' cos.  &'  /riCv 

Let  cos.  w  as  =r r,  (545) 

2  cos.  a  cos.  6 

and  we  have,  by  (544), 

cos.E/-\-cos.(a,-^b,)  =  2cos.mcos.E-^2cos.mco3.(a  +  b) 
z=zcos.(E+m)+cos.(E— m)+cos.(a+6+m)+cos.(a+6— m) 
cos.  JEJ'  as  —  cos.  (a'  +  6')  +  cos.  (E  +  m)  +  cos.  ( JS— m) 

+  cos.  (a  +  5  +  m)  +  cos.  (a +  6  —  w),   (546) 


316  SPHERICAL    ASTR0N0M3T.  [CH.  VIII. 

By  a  lunar  distance. 

whence  E'  can  be  found  by  a  table  of  natural  sines  and  co- 
sines, when  m  has  been  found  from  (545). 

II.  In  the  same  way  by  (279),  we  find 

2(sin.iZ)^C-^=^^=C-2!l^^)-^'    (547) 
cos.  a  cos.  o  cos.  a'  cos.  b' 

cos.  (af  —  b')  —  cos.  E'  =  2  cos.  m  cos.  (a  —  b) — 2  cos.  m  cos.  E 
z=z  cos.  (a — b-\-m)-\-cos.a — b — m) — cos.(E-\-m) — cos.  (22 — m) 
cos.  E1  =.  cos.  (af — b')  —  cos.  {a — b-\-m) — cos. (a — b — m) 

+  cos.  (E  +  m)  +  cos.  (E—m).      (548) 

III.  The  correction  may  be  separated  into  two  parts,  one  of 
which  depends  only  upon  the  sun  or  star,  and  the  other  upon 
the  moon ;  and  let 

d'  E  =  the  part  of  <?  E  which  depends  upon  the  sun  or  star, 
d"E  5=  the  part  which  depends  upon  the  moon. 

Now  if  the  correction  were  only  to  be  made  for  the  moon, 
SM  would  be  decreased  to  SM1,  whence 

SM'  =  E-\-d"E, 
and  if  we  put 

S=ZSMt     M—ZMSy 
s  =  i  (a  +  t>  +  E)>  (549) 

the  triangles  SMM1  and  SZM  give 

(sin.^M)^sin'Ssin^5-=^l 
sin.  E  cos.  6 

_  *in.[E  +  i(3»E  —  *b)]  sm.j(S"E  +  S b) 
sin.  d  b  sin.  E 


$  103.]  LONGITUDE.  317 

By  a  lunar  distance. 

<S"  E  -4-  <*  b 

[1  +  J  cotan.  E  sin.  1".  (*"  JS— <*&)]   (550) 


r   2^6 

60^,^^(59-4^-^)  +  sin-5sin>(^-"a)  .  -*!£ 
v  '    '  sin.  £  cos.  6 

+  t18"  —  i  cotan-  *  sin-  1"  [(*"  E)2  —  (d  &)*]•    (551) 
The  triangles  SS'M  and  SZM  give,  by  (277)  and  (281), 

(cos.  iS)2  =  ^(s-^ls-a) 
v  '  sin.  E  cos.  a 

_  sin.  [JE  +  j  (J"  JE  —  8  a)]  sin,  £  (<?"  .E  -f-  j  a) 
sin.  d  a  sin.  JS 

60'+  rig  =  (W-  *«)  +  ^aUf;>  Sjn^-a) . ^.    (553) 
v  '  '     -  sm.  E  cos.a     v       ' 

If  now  M1  K  and  S'L  are  drawn   perpendicular  to  MS, 
and  £'£/  to  JfcT'jSf,  we  have  nearly 

S'M'=  E  +  9  E  =  SM'  +  SL'  —  E  +  6"E  +  SL' 

'  9E=9"E  +  SL'  =  9»E  +  9'E  +  (SL,—lfE)  (554) 
§'E=.SL  =  da  cos.  #  (555) 

SL'z=*a  cos.  (£'££')  =£  3  a  cos.  (S—  MSM') 
=  8  a  cos.  S  +  ^  a  sin.  #  sin.  JBS4P 
=  *'E  +  S'L  sin.  JHSflf'  (556) 

#£'  —  d'E  =  S'L  sin.  jff&Jf7.  (557) 

But  from  MSK, 

sm.  MSM1  =  —. — =-  =     — — = — ,  (558) 

sin.  E  sm.  E 

27* 


318  SPHERICAL    ASTRONOMY.  [CH.  VIII. 

Lunars.  Tables  XVII,  XVIII,  XIX,  XX. 

whence 

SL'-*E  =  8-'LXM>K„S'mA-,  (559) 

sin.  E  ' 

and 

ifr^*g+Vj+*A  ***'**  (560) 

1  '  sin.  E  v       ' 

2°  +  m  =  (60'  -f  *  JEJ)+(60'+*//JEJ)+  gin  js  (561 ) 

in  which  1°  is  added  to  <S'  JE?  and  <$"  JEJ  in  order  to  render  them 
positive.  Now,  of  60'  +  9' E  (553),  the  part  60'  —  6  a  is 
given  in  table  XVII  or  table  XVIII ;  and  the  remaining  term  is 
computed  by  proportional  logarithms,  and  is  the  first  correction 
of  the  First  Method  of  the  Navigator.  [B.  p.  231.]  The  pro- 
portional logarithm  of  the  factor  2  3  a  sec.  a,  is  the  logarithm 
of  the  table  from  which  60'  —  §a  is  taken. 

In  the  same  way,  the  two  first  terms  of  60'  -f-  d"  E  are  taken 
from  table  XIX  and  (551).  The  remainder  of  (551)  com- 
bined with  the  third  term  of  (561),  is  computed  and  inserted 
in  table  XX  of  the  Navigator. 

In  calculating  table  XX,  the  value  of  &"  E  is  used,  which 
is  obtained  from  the  two  first  terms  of  (551);  and  S'L  and 
M'K  are  found  from  S'SL  and  MKM1  in  which  the  sides 
are  so  small  that  their  curvature  may  be  neglected,  and  we 
have,  nearly 

'8'L  =z  */{8  «2  _  */ E2)  (562) 

M'K=*/{&b*—j'E*).  (563) 

IV.  The  calculation  of  the  values  of  d  a  and  db  will  be 
fully  explained  in  subsequent  chapters ;  but  we  need  only  re- 
mark, in  this  place^  that  the  value  of  <j  a  for  a  star  is  given  in 
table  XII,  for  the  sun  it  is  the  number  of  table  XII  diminished 


<$>  103.]  LONGITUDE.  319 

By  a  lunar  distance. 

by  that  of  table  XIV,  and  for  a  planet,  it  is  that  of  table  XII, 
diminished  by  that  of  table  X,  A.  The  value  of  8  b  is  obtained 
by  the  formula 

H  —  P  cos.  b  —  8'b9  (564) 

in  which  81  b  is  the  number  of  table  XII,  and  P  is  the  number 
taken  from  the  Nautical  Almanac,  and  which  is  called  the 
horizontal  parallax.  In  computing  table  XX,  the  value  of  P 
is  taken  at  its  mean  of  57'  30". 

In  the  formulas  for  the  corrections,  the  zenith  distances  may 
be  introduced  instead  of  the  altitudes,  and  if  we  put 

90°  —  a  =  Z,     90°  —  b  =  Z, 

s1=zz  +  Z+  E,  (565) 

we  have,  by  neglecting  the  term  depending  upon  the  correc- 
tion of  table  XX,  as  well  as  the  other  small  quantities, 

sin.  5  j  sin.  (s1  —  z) 


cos.^iM  — 


sin.  E  sin.  Z 


_  sin.  [E  +  j  (8"E  +  9  b)]  sin.  £  (8  b  —  9" E) 
sin.  E  sin.  8  b 


9  b  —  8"  E 

„  „  .         2  sin.  s,  sin.  (5,  —  z)     , 

d"E  =  8b •-,-», ~  *  & 

sin.  E  sin.  Z 

sin.Sj  sin.  (*.  —  Z)        JE  +  da 

cos.2  J  8  =  ? — =V L  =  — j-j 

sin.  E  sin.  z  2  8  a 


(566) 
(567) 


2  sin.  s.  sin.  (s.  —  Z)   . 
r — ^  \  * <*a. 


^  =  ~^-| '  .*    ^  v,  1 L  8  a.  (568) 

sin.  E  sin.  2  ' 

Then  the  second  term  of  the  value  of  8'  E  is  the  first  cor- 
rection of  the  Third  Method  of  the  Navigator  [B.  p.  242.] 


320  SPHERICAL   ASTRONOMY.  [CH.  VIII, 

By  a  lunar  distance. 

and  the  second  term  of  the  value  of  $"  E  is  the  second  cor- 
rection of  this  method  ;  and  the  computation  from  (564,  567, 
568)  agrees  entirely  with  this  method.  The  third  correction 
is  taken  from  table  XX,  as  in  the  first  method. 

V.  Draw  ZN  perpendicular  to  MS,   so   as  to   make   SN 
acute.     In  the  right  triangles  ZSN  and  ZSM  let 

B=z90°  —  SN,  B'z=z90°+MN,  A  =  i(B'  +  B)f  (569) 
and  we  have 

E  ==  3IN+  SN=B'  —  B,  (570) 

and,  by  Bowditch's  Rules  for  oblique  triangles, 

cos.  ZS  :  cos.  ZM  ==  cos.  NS  :  cos.  MN, 
or  sin  a  :  sin.  b  =  sin.  B  :  sin.  B ;  (571 ) 

and,  by  the  theory  of  proportions, 

sin.  a  -\-  sin.  b sin.  B  -f-  sin.  B 

sin.  b  —  sin.  a        sin.  B1  —  sin.  B' 

that  is, 

tang.  J  (a  -f-  b)  _     tang.  A 

tang.  £  (6  —  a)  ~~  tang.  £  E 
tang.  A  =  tang.  £  (a  -j-  b)  cotan.  J-  (b  —  a)  tang.  J  E  (573) 
B'  =  A  +  iE,     B  —  A  —  ^E,  (574) 

and  the  right  triangles  Z#iV,  MZN,  SLS1,  MKM',  give 

cos.  £  =s =  cotan.  ZS  tang.  $ZV  =  tang,  a  cotan.  2? 

d  CI 

— cos.  ilf=  -Tjr  —  —  cot«  ZMtmg,  MN  =  tang.  &  cotan.  i?' 

#  E  =  da  tang,  a  cotan.  JB  (^75) 

<S".E  =s  <*  6  tang,  b  cotan.  JB',  (576) 


(572) 


<§>   103.]  LONGITUDE.  321 

Lunars.  Table  XLVII. 

and  the  formulas  (573-576)  correspond  to  the  Fourth  Method 
of  the  Navigator.  [B.  p.  243.] 

It  may  be  observed,  that  since  cotan.  J  (6  —  a)  is  the  only 
term  of  (573)  which  can  change  its  sign,  A  is  acute  when 
b  is  greater  than  a,  and  obtuse  when  b  is  less  than  a. 

VI.  The  most  important  of  corrections  of  the  distance  arise 
from  that  term  of  8  b  (564),  which  depends  upon  the  parallax. 
If  we  consider  this,  therefore,  as  the  only  correction  of  the 
moon's  altitude,  we  may  calculate  the  corrections  of  the  dis- 
tance arising  from  it  by  putting 

9  b  =z  MM>  =  P  cos.  b.  (577) 

The  triangles  ZSM  and  M'MK,  give  then 

&'E  sin.  a  —  cos.  E  sin.  b        ,„m~K 

cos.  M=z  — -  == : — - - (578) 

x*cos.  6  sin.  E  cos.  6 

d"E  s=  —  P  sin.  a  cosec.  E  +  P  cotan.  E  sin.  b,  (579) 

and  if  we  put 

^E'  =  P  sin.  a  cosec.  E  (580) 

S2E  —  ±P  cotan.  E  sin.  b,  (581) 

in  which  the  signs  are  taken  so  that  <?2  E  is  always  positive, 
we  have 

d"E=.  —  \E±*2E  (582) 

10°  +  8"E  —  (5°  —  s±  E)  +  (5°  ±  * 2  E).       (583) 

Now  table  XLVII  is  a  common  table  of  proportional  loga- 
rithms, like  table  XXII ;  but  the  angle  which  is  placed  at  the 
top  of  the  table  is 

5°  —  the  angle  of  Table  XXII,  (584) 


322  SPHERICAL    ASTRONOMY.  [CH.  VIII. 

Lunars.  Tables  XLVIII,  XLIX,  L. 

and  the  angle  at  the  bottom  of  the  table  is 

5°  +  the  angle  of  Table  XXII ;  (585) 

so  that  the  terms  of  (583)  may  be  directly  obtained  from  these 
tables ;  and  this  method  of  computing  the  corrections,  which 
depend  upon  the  moon's  parallax,  agrees  with  the  second 
method  of  the  Navigator.   [B.  p.  239.] 

The  remaining  corrections  may  be  computed  from  the  for- 
mulas (567  and  568),  and  the  corrections  of  table  XX  may 
be  neglected,  provided  the  value  of  E  is  corrected  for  the 
parallax.  These  combined  corrections  may  be  inserted  in  a 
table  like  table  XLVIII,  which  serves  for  the  star,  and,  by 
means  of  the  part  P,  for  the  sun ;  or  like  tables  XLIX  and 
L,  which  serve  for  the  planets.  In  calculating  those  tables, 
the  moon's  horizontal  parallax  is  taken  at  its  mean  value  of 
57'  30" ;  and  the  planet's  or  sun's  parallax  in  altitude  is  ob- 
tained from  the  formula 

<*'  a  s±  —  P  cos.  a, 

in  which  P  is  the  horizontal  parallax.  The  value  of  P,  used 
in  the  construction  of  the  part  P  of  table  XLVIII,  is  8".6 ; 
that  used  for  table  XLIX  is  35" ;  and  since  these  corrections 
are  proportional  to  the  parallax,  they  are  easily  reduced  to  any 
other  parallax.     This  reduction  is  actually  made  in  table  L. 

VII.  The  value  of  d"  E  (578),  might  be  found  by  the  for- 
mula 

8„E  __  2  sin,  a  —  sin,(6  +  jE:)  —  sin.  (6—  E)  ' 

2  sin.  E  ? 

which  is  easily  calculated  by  means  of  the  table  of  natural 
sines  and  cosines. 


§  103.]  LONGITUDE.  323 

Lunars.  Distance  from  Nautical  Almanac. 

VIII.  The  true  distance  may  be  obtained  from  observation 
by  either  of  the  preceding  methods,  and  the  time  of  the  ob- 
servation must  be  compared  with  the  time  when  the  distance 
is  the  same  to  an  observer  at  Greenwich.  Now  this  latter  time 
can  be  obtained  from  the  Nautical  Almanac  by  precisely  the 
same  process  of  interpolation,  which  has  been  applied  to  the 
changes  of  right  ascension.  The  distances  are  given  in  the 
Nautical  Almanac  for  every  three  hours,  and  the  proportional 
logarithm  of  the  difference  of  these  distances.  If,  then,  the 
distance  increases  uniformly  at  the  rate  of  increase,  F  for 
every  three  hours ;  the  interval  T,  at  which  it  has  increased 
by  the  quantity  F1,  is  found  by  the  proportion 

F:  F'  —  Sh  :  T  (587) 

Prop.log.  T—  Prop.  log. jF'  — Prop,  log. F+  Prop. log. 3\  (5S8) 

But  Prop.  log.  3*  =  0 ;  (589) 

and  if  we  put 

Prop.  log.  F~Q,  (590) 

(588)  becomes 

Prop.  log.  T  ±=  Prop.  log.  F<  —  Q.  (591) 

If  the  distance  increased  uniformly,  the  value  of  Q  would 
be  invariable  ;  but  Q  is  variable,  and  must  be  regarded  as 
belonging  to  the  middle  instant  of  the  interval  to  which  it  be- 
longs ;  and  it  increases  while  the  distance  decreases,  and  the 
reverse.     Let  then 

<5  Q  z=  the  decrease  of  Q  in  three  hours, 

<*  T  =  the  correction  of  T,  arising  from  the  change  of  Q, 

and  the  value  of  Q  for  the  interval  T  is 

Q+ 180^- ^  «  =  ^  +  *'«>  <592> 


324  SPHERICAL    ASTRONOMY.  [CH.  VIII. 


Lunar  distance  from  Nautical  Almanac. 

so  that  by  (591)  and  (340) 

Prop.  log.  (T+aT)  =  Prop.  log.  T  —  *  Q 

(593) 

log.(T+3T)  =  \og.  T+6'Q 

(594) 

.    .  ~,v        ,        m      ,       /,*         *Z\        „  _ 

log. 

But  if  in  (196)  and  (204)  we  substitute 

S  T 

-jr^Z™,  (596) 

we  have,  by  logarithms  and  (595), 

;  log-  =  /f>g.  (l+^)  -  ^,  (597) 

so  that  by  (592)  and  (208) 

a  T  -    ^Q  -=    (18Qfft-r)  ^Q  /iboov 

*~    log.  6  2X  180 -.X  0.434  V       ; 


(180OT—  T)  TSQ 
:  156  OT 


(599) 


and  the  table  [B.  p.  245.]  for  correcting  by  second  differences 
may  be  calculated  by  this  formula ;  and,  in  order  to  obtain  the 
value  of  9  T  expressed  in  seconds,  the  factor  T  should  be 
expressed  in  seconds,  while  (180771 —  T)  is  expressed  in 
minutes ;  and  it  must  not  be  forgotten,  that  the  proportional 
logarithms  are  decimals. 

IX.  When  the  distance  is  observed  for  a  star,  whose  dis- 
tance is  not  given  in  the  Nautical  Almanac,  the  Greenwich 
time  of  the  observation  can  be  found  approximately  by  adding 
the  assumed  longitude,  if  west  to  the  observed  time,  or  subtract- 
ing it  if  east ;  or  the  time  can  be  taken  from  the  chronometer 
if  it  is  regulated  to  Greenwich  time. 


<§>   103.]  LONGITUDE.  325 


Lunar  distance  not  given  in  the  Nautical  Almanac. 


Find,  in  the  Nautical  Almanac,  the  right  ascension  and 
declination  of  the  star  and  the  declination  of  the  moon,  for 
this  time.  Then,  if  T  and  S  (fig.  43.)  are  supposed  to  be  the 
moon  and  star,  and  P  the  pole  of  the  equator,  D  and  D'  their 
declinations,  disregarding  their  names,  so  that  their  polar  dis- 
tances are  90°  ±  D  and  90°  ±  D1,  and  if  R'  is  their  differ- 
ence of  right  ascensions,  we  have,  when  their  declinations  are 
of  the  same  name,  by  putting 

8=  i(D  +  D'  +  E)  (600) 

cos^^  =  cos.^P^v(-^?"(^V    (601) 
2  *  ^  \    cos.  D  cos.  D1     J     v       ' 

But  if  the  declinations  are  of  the  same  name, 

^  \    cos.  D  cos.  D'    / '    v       ' 

and  the  right  ascension  of  the  moon  being  thus  found,  the 
Greenwich  time,  when  it  has  this  right  ascension,  is  easily 
found  from  the  moon's  hourly  ephemeris  in  the  Nautical  Al- 
manac, and  this  method  is  the  same  with  that  in  [B.  p.  428.] 

X.  The  latitudes  and  longitudes  may  be  used  instead  of  the 
right  ascensions  and  declinations,  and  the  calculation  will  be 
as  in  [B.  p.  427.]  The  variation  of  daily  motion  is,  in  this 
case,  to  be  had  regard  to,  precisely  as  explained  in  (536-541). 

XI.  The  distances  of  the  Nautical  Almanac  can  be  calcu- 
lated from  the  right  ascensions  and  declinations  of  the  sun, 
moon,  and  stars,  or  their  latitudes  and  longitudes,  by  resolving 
the  triangles  TPS  (fig.  43.)  by  either  of  the  methods  which 
have  been  given,  when  two  sides  and  the  included  angle  are 
known,  as  in  [B.  p.  434.] 

28 


326  SPHERICAL    ASTRONOMY.  [CH.  VIIL 

^  Lunar  distances. 

In  calculating  the  distance  of  the  sun  and  moon,  the  latitude 
of  the  sun  may  be  usually  neglected ;  so  that  if  SR  (fig.  46.) 
is  an  arc  of  the  ecliptic,  S  the  sun's  place,  M  the  moon's, 
and  MR  perpendicular  to  SR, 

MR  z=z  L  —  the  moon's  latitude, 

SR  z=  Lx  =-  the  difF.  of  long,  of  ©  and  J) , 

and  cos.  E  =  cos.  SM  —  cos.  L  cos.  Zx,  (603) 

as  in  [B.  p.  433.] 

It  would,  however,  be  rather  more  accurate  to  take 

L  z=  the  diff.  of  lat.  of  ©  and  J) . 

XII.  The  determination  of  the  longitude  by  solar  eclipses 
and  occultations  will  be  reserved  for  another  chapter. 

104.    Examples. 

1.  Calculate  the  correction  of  table  XLV,  when 

T=zlh  50w,  and  B  =  9m  =  540s. 

Solution.  lh  50m  P.  L.  lm  50*  ar.  co.  8.0080 

12*  —  1*  50"  =  10*  10"  P.  L.  10*  10*  ar.  co.  8.7519 

2  P.  L.  12m  2.3522 

£  B  =  270'  2.4314 


corr.  =  34ff.9  1.5435 

2.  Calculate  the  correction  of  table  XLV,  when 
T=3*  10",  and  B  =  IP*. 

Ans.    64M, 


§   104.]  LONGITUDE.  327 

To  find  the  moon's  right  ascension. 


3.  Find  the  right  ascension  of  the  moon's  bright  limb,  Sept. 
25,  1839,  at  the  time  of  the  transit  over  the  meridian  of  New 
York.  The  right  ascensions  of  the  moon  for  the  two  preceding 
and  the  two  following  transits  at  Greenwich  are 

Sept.  25.  Moon  II.  L.  T.  2*  0m  36*.69 
Moon  II.  U.  T.  2  30    38  .08 

Sept.  26.  Moon  II.  L.  T.  3  1  33  .18 
Moon  II.  U.  T.    3  33    19  .89 

Ans.     2M3W  14».4. 

4.  At  a  place  in  west  longitude,  Oct.  25,  1839,  the  moon's 
bright  limb  passed  the  meridian  10m  6'.83  sideral  time,  before 
the  star  C.  Tauri  ;  find  the  longitude  of  the  place  of  observa- 
tion. 

The  fight  ascension  of  the  star  C.  Tauri  was  5*43m  16'.84, 
and  those  of  the  moon 

Oct.  25.  Moon  II.  L.  T.  4*43*53 '.55 
Moon  II.  U.  T.    5    16    28  .40 

Oct.  26.  Moon  II.  L.  T.  5  52  51  .91 
Moon  II.  U.  T.    6  26    40  .00 

Ans.     70°  25'  30"  W. 

5.  Find  the  moon's  parallax  in  altitude,  and  the  correction 
and  logarithm  of  table  XIX,  when  the  altitude  is  40°  40',  and 
the  horizontal  parallax  is  58'. 


328  SPHERICAL    ASTRONOMY.  [CH.  VIII. 

"~  Tables  XVII,  XVIII,  XIX,  XX. 


Solution.  58'  P.  L.  0.4918 

40°  40'  sec.  0.1200 


Parallax  in  alt.  =.         44'  P.  L.  0.6118 


By  Table  XII.     Refrac.  =  V  6"  9.6990 

Corr.  —  16'  48"*  =  59'  42"  —  42'  54"     P.  L.  0.6228 


Log.  of  Table  XIX.     =     0.2018 

6.  Find  the  correction  and  logarithm  of  Table  XVII,  for  a 
star,  when  the  altitude  is  13°.  15'. 

Ans.     Corr.  =  5G'  2",  Log.  —  1.3433. 

7.  Find  the  correction'  and  logarithm  of  Table  XVII,  for 
Venus  or  Mars,  when  the  parallax  is  20",  and  the  altitude 
24°  30. 

Ans.     Corr.  =  58/  14",  Log.  =z  1.6647. 

8.  Find  the  correction  and  logarithm  of  Table  XVIII, 
when  the  altitude  is  56°. 

Ans.     Corr.  —  59'  26",  Log.  —  1.9544. 

9.  Find  the  correction  and  logarithm  of  Table  XIX,  when 
the  altitude  is  70°,  and  the  horizontal  parallax  54'. 

Ans.     Corr.  =  41'  34",  Log.  =  0.2299. 

*  The  numbers  of  Table  XIX  are  so  disposed  in  the  Navigator,  that 
the  corrections  of  proportional  parts  of  parallax  are  all  additive.  This 
is  effected  by  placing  each  number  opposite  that  parallax,  which  is  10" 
less  than  the  one  to  which  it  belongs.  There  is,  therefore,  a  correction 
for  0"  of  parallax. 


§  104.]  LONGITUDE.  329 

Auxiliary  angle  in  lunar  distances. 


10.  Compute  the  value  of  the  auxiliary  angle  m,  in  the  first 
and  second  methods  of  correcting  the  lunar  distance,  when 
the  moon's  apparent  altitude  is  40°  40',  its  horizontal  parallax 
58',  and  the  sun's  apparent  altitude  70°. 

Solution.  The  values  of  m  might  be  computed  directly  from 
(545),  but  it  is  more  convenient  to  obtain  it  by  some  process 
of  approximation.     For  this  purpose  let 

m  =-  60°fc+  8  m, 
and  we  have 

2  cos.  (60°  +  »  .)  =  «»(»  +  ' »)co..(«- J  a) 
v         '        '  cos.  o  cos.  a 

=*s  2  cos.  60°  cos.  9  m  _  2  sin.  60°  sin.  §  m  (604) 

=  (cos.  8b — tang,  b  sin.  8  b)  (cos.  ^a-j-tang.  a  sin.  8a), 

in  which  we  may  put 

2 cos. 60°  =  1,  cos. <? 6  =  1—  2sin.2£tf&=l—  £<5&2sin.2l" 

cos.  8 m=zl  —  \8m?  sin.2  I", 

and  (604)  becomes 

2  8  m  sin.  60°  =  8  b  tang,  b  —  8  a  tang,  a  (605) 

+  %(  8  b2— 8m2)  sin.  1". 

But  if  we  take 

e=.2  8  b  sec.  b  and  e'  =  2  d  a  sec.  a, 

Prop.  log.  e  is  the  logarithm  of  Table  XIX,  and  Prog.  log.  t'  is 
the  corresponding  logarithm  for  the  sun,  star,  or  planet,  and 
by  (605), 

8  m  =  I  e  sin.  b  cosec.  60°  —  |  ef  sin.  a  cosec.  60°     (606) 
.{_£(  $b2  —8m2)  sin.  1"  cotan.  60% 
28* 


330  SPHERICAL   ASTRONOMY.  [CH.  Till. 

Auxiliary  angle  in  lunar  distances. 


whence  in  the  present  case 

c              P.L.     0.2018  e'         P.  L.     2.0173 

40°  40'         cosec.     0.1860  70°      cosec.     0.0270 

60°                 sin.     9.9375  9.9375 


1°25'7"                   0.3253  153"            1.9818 
approx.  3m=l(  1°  25'  1"— V  53")=£(  1°  23'  4")=20'  46"=  1246" 
I  b  as  42'  54"  as  2574" 

db+  dm  =  3820  3.5821 

H—  dm=z  1328  3.1232 

1"  sin.     4.6856 

60°  cotan.     9.7614 


corr.^m  as  7"  as  J  (14")  1.1523 

I  m  as  20'  46"  +  7"  as  20'  53". 

11.  Compute  the  value  of  the  auxiliary  angle  m,  when  the 
moon's  apparent  altitude  is  25°  30',  the  horizontal  parallax  60', 
and  the  star's  apparent  altitude  10.° 

Ans.     60°  13'  48". 

12.  Find  the  correction  of  Table  XX,  when  the  distance 
is  25°,  the  sun's  altitude  10°,  and  the  moon's  altitude  25°. 


§   104.]  LONGITUDE.  321 

Table  XX. 

Solution.  We  should  find,  in  this  case, 

db  =  50'  6"  *a  —         5'   6" 

9"E  =  —  27'  22"  y.E  =    —  3'  15" 


9b—*"E=  1°  17' 28"  =4648"  <?  a  —  *'JE  =         8' 21"  =501" 

a6  +  ^£  =  22/44//  *a+3'.E  =         1'51"  =111" 

22/  44"  s-  P.  L.  0.8986  0.899 

1°  17'  28"  =  4648"  (ar.  co.)  6.3327  P.  L.  0.366 

25°  tang.  9.6687  2  sin.  9.252 

1"  cosec.  5.3144  l"2cosec.  0.629 


V  6"  =  66"  2.2144      501"  (ar.  co.)  7.300 

i  (66")  =  33"  111"  (ar.  co.)  7.955 

2)6.401 

24"  =  18"  +  6"  3.200 

57"  =  corr.  Table  XX. 

13.  Calculate  the  correction  of  Table  XX,  when  the  dis- 
tance is  120°,  the   sun's  altitude  20°,   and  the  moon's  altitude 

10°. 

Ans.     10". 

14.  Calculate  the  corrections  of  Tables  XLVIII,  XLIX, 
and  L,  when  the  apparent  distance  is  28°,  the  moon's  apparent 
altitude  38°,  the  planet's  apparent  altitude  18°,  and  its  hori- 
zontal parallax  16" 


332  SPHERICAL    ASTRONOMY.  [CH.  VIII. 


Tables  XLVIII,  XLIX,  L. 


Solution. 

57'  30"  P.  L.  0.4956  0.4956 

18°  cosec.  0.5100  38°    cosec.  0.2107 

28°  sin.  9.6716  tang.  9.7257 


5°— lst.cor.=4°22'9"    0.6772  50+2dcor.z=6°6'34"  0.4320 
6°  6'  34"  moon's  par.  in  alt.  =  43' 

28°  moon's  approx.  alt.  ==  38°  43' 


28°  29'  ==  approx.  dist. 
18°  43'+29'=72':=4320"ar.co.  6.3645 

38°  43'  43—  29=:14'  P.L.  1.1091 


28°  22'  =  i  sum  tang.  9.73235  28°     tang.  9.7257 

10°22'  =  Jdiff.        cotan.  0.73771  1"     cosec.  5.3144 


£(28°)  =  14°  tang.  9.39677  2)34"  2.5137 

A=z36°2V  9.86683  17" 

lstang.=:22021'        tang.  9.6140  9  614 

18°  cotan.  0.4882  0.488 

By  Table  XII  2'  54"     P.  L.  1 .7929     T.  X,  A.  33",  P.  L.  2.515 

2'  17"  1.8951     25"=  cor.  T.  XIX  2.617 

2d  ang.  =  50°  21'         tang.  0.0816    ^fX25"=ll"=cor.T.L 

38°  cotan.  0.1072 

By  Table  XII  1'  13"    P.  L.  2.1701 

47"  2.3589 

Cor.  Table  XLVIII  =  2"  17"  —  47'  +  17"  =  V  47". 


§  104]  LONGITUDE.  333 


True  from  apparent  distance. 


15.  Calculate  the  corrections  of  Tables  XLVIII,  XLIX, 
and  L,  when  the  apparent  distance  is  60°,  the  moon's  appa- 
rent altitude  59°,  the  planet's  apparent  altitude  30°,  and  its 
horizontal  parallax  30." 

Ans.     Cor.  Table  XLVIII  =  V  24" 

XLIX     =  —  21" 

L         =  —  18". 

16.  Find  the  correction  of  the  table  [B.  p.  245.]  for  the 
interval  of  2h  30 m,  and  the  difference  of  the  Proportional  Log- 
arithms equal  to  83. 

Ans.     15  s. 

17.  If  the  observed  distance  were  45°  34'  10",  the  moon's 
apparent  altitude  22°  19',  its  horizontal  parallax  60'  19",  the 
planet's  apparent  altitude  42°  12',  its  horizontal  parallax  15".3 ; 
what  is  the  true  distance. 

Solution.     I.      In  this  case  m  =z  60°  12'  28" 

«—  42°  12'  a  a  =  51"        a=42  11    9 

6  =  22  19  6b=z  53'  31"  &'  — 23  12  31 

«'+&'  =  65  23  40"—  N.cos.=—  0.41637  E=z  45  34  10 
E-{-m=\05  46  38       N.cos.=— 0.27189 
a-f-6+m  =  124  43  28      N. cos.=—  0.56964 


—  1.25790 

E— m=— 14° 38 '18"  N.  cos.  =  0.96754 
a  +  b  —  m=z       4  18  32  N.  cos.  =  0.99718 


E'  =  45  121  N.  cos.  =  0.70682 


334  SPHERICAL    ASTRONOMY.  [CH.  VIII. 


Lunar  distance  corrected. 


II.  a  —  b  +  m=:      80°  5' 28"  —  N.  cos.  =  —  0.17208 

a  —  b  —  m  = —40  19  28   —  N.  cos.  =  —  0.76239 

E  +  m  —    105  46  38     N.  cos.  as  —  0.27189 


—  1.20636 

a!  —  b<  =      18°  68'  38"         N.  cos.  =      0.94567 
E  —  m  —  — 14  38  18  N.  cos.  —      0.96754 


£'  „*=      45     1  14  N.  cos.  —      0.70685 

III.       s  =  J  (a  +  b  +  IB)  =  55°  2'  35"         sec.  0.2420 

JE  =  45°34  10"       sin.  9.8538  9.8538 

s  —  a  =  12  50  35    cosec.  0.6532  0.6532 

s—E  —    9  28  25       sec.  0.0060     611".  T.  XIX.  0.1920 


598".  Table  XVII  1.8907  20  37  P.  L.  0.9410 


43"  P.  L.  2.4037        31     Table  XX. 


59°.51  27'  19" 

E  '  =  45°  34' 10"  +  59/  51"  +  27'  19"  —  2°  =  45°  1  20". 


$104.]  LONGITUDE.  335 


Clearing  lunar  distances. 


IV.  Z  =z  47°  48'  z  =  67°  41' 

51=80°31/35//  cosec.  0.0060 
E=z45  34  10      sin.  9.8538 
9.6990 


9.5588 

9.5588 

Z=47°48/           sin.  9.8697 

s=67°41'     sin.  9.9662 

^—2=12°  50' 35"  cosec.  0.6532 

sx  —  Zz=32°43'35" 

cosec.  0.2671 

i  a  =  51"             P.  L.  2.3259 

a  6  =53' 21"  P.  L.  0.5281 

1st  cor.  ==  42"             P.  L.  2.4076  2d  cor.  1°  26'  7"          0.3202 

<>&  =  53'2i" 

a  a  =  •     51" 

£'=54'    3"+45°34'10"+31". 

-18"-1026'58"=4501'28" 

V.         J  (a +  6)  =  32°  15' 30" 

tang.  9.80014 

J(a_&)  —    9  56  30 

cotan.  0.75626 

IB          =  22  47    5 

tang.  9.62330 

^1           =123  28  15 

tang.  0.17970 

lstang.  ==  100°  51'  10"  tang.  0.7175 

2d  ang.  =  146°  15' 20"  tang.  9.8250 

a    —   42°12/       cotan.  0.0425   6z=22°  19;  cotan.  0.3867 

d  a    =  51"  P.  L.  2.3259  db=.  5321"  P.L.  0.5281 


1st  cor.  =  —  7"  P.L.  3.1859 

2d  cor.  =  —  32'  46"  P.  L.  0.7398 

E'  ss  45°  34'  10"  —  7" — 32'  46" + 31 "  — 18"  =  45°  V  20". 


336  SPHERICAL    ASTRONOMY".  [CH.   VIII. 


Clearing  lunar  distances. 


VI.  60'  19"  P.  L.  0.4748  0.4748 

a  ==  42°  12'        cosec.  0.1728  b  ~  22°  19'  cosec.  0.4205 
E  ==  45°  34'  10"      sin.  9.8538  tang.  0.0086 


1st  cor.  —    4°    316"  0.5014  2d  cor.  zz:5°  2228"  0.9039 

Cor.  Table  XLVIII  =  1'  33" 
JE  =45°  34' 10"4-4°  3' 16"+5°  22' 28"+!'  33"— 10°— 45°  V  27". 

VII. 

a—     42°12/    N.sin.   0.67172 

b  +  E  =  67°  53' 1 0",  AN.  sin.—  0.46322  60'  19"  P.  L.  0.4748 

b  —  E  ——23°  15'  10",  £N.  sin.   0.19739 


0.40589       ar.  co.       0.3916 
£  =  45°  34  10"  sin.  9.8538 


Cor.  Table  XLVIII  =  V  33"  cor.zz:  —  34' 17"  0.7302 

E  =  45°  34'  10"  +  T  33"  —  34'  17"  =  45°  1'  26". 

18.  The  apparent  distance  of  the  sun  and  moon  is  95°  50' 
33"  ;  the  moon's  apparent  altitude  is  35°  45'  4",  its  horizontal 
parallax  is  54/24//;  the  sun's  apparent  altitude  is  70°  48'  1"  ; 
what  is  the  true  distance  ? 

Ans.     95°  44'  29". 

19.  The  apparent  distance  of  a  star  from  the  moon  is  31°  13' 
26" ;  the  moon's  apparent  altitude  is  8°  26'  13",  its  horizontal 
parallax  is  60',  the  star's  apparent  altitude  is  35°  40' ;  what  is 
the  true  distance  ? 

Ans.     30°  23'  ^Q". 


§   104.]  LONGITUDE.  337 

Lunar  distances. 


20.    Find  the  Greenwich  time,  Oct.  3,    1839,   when  the 
moon's  distance  from  the  sun  was  38°  12'  9". 

Solution. 
Distance  1839,  Oct.  3,  15*  38°  59' 2 1"     P.  L.  0.3180 

38  12    9 


18*  P.  L.  3189  47' 12"     P.  L.  0.5813 


3180         T=    1*38™  10*  P.L.  0  2633 


9  cor.T.zn  —2* 


Greenwich  time  =    16*  38™   8*. 

21.  Find  the  Greenwich  time,  Jan.  2,  1839,  when  the  moon's 
distance  from  Aldebaran  was  70°  45'  13". 

1839.  Jan.  2,  9*  Greenw.  Time,         Dist.  —  69°  26'  29" 

P.  L.  =  0.2852 
12*  P.  L.  —  0.2863 

Ans.     12*  31™  47* 

22.  The  correct  distance  of  the  moon  from  /*Corvi,  1839, 
April  3d,  11*  20™,  in  longitude  70°  W  by  account,  was  54° 
8'  15" ;  what  was  the  longitude  ? 


29 


338  SPHERICAL   ASTRONOMY.  [CH.  VIII. 


Lunar  distances. 


Solution.  54°   8' 15"    Gr.T.=  11*  20™  +  4*  40™  =:  16* 

3>'s  Dec.  s±  26  48  52       by  N.  A.     sec.     0.04941 
*'s  Dec.  =  22  30  11  sec.     0.03439 


\  sum        =  51°  27'  18"  cos.     9.79198 

Dist.—  J  sum  =    2  24  36  cos.     9.99962 


2)19.87540 


3*  59 m  42*  cos.     9.93770 

♦  's  Dec.  =  12   25    56 


3>'s  Dec.  =  16*  25m  38*  r=  Greenw.  Time  ==  16* 
Long.  =  16*—  14*  20™  =  4*40m  =  70°,  as  supposed. 

23.  The  correct  distance  of  the  moon  from  Castor,  1839, 
Nov.  29 d  19*,  in  longitude  45°  W.  by  account,  was  78°  3'; 
what  was  the  longitude  1 

Greenwich,  1839, 
Nov.  29*  21*,   D  's  R.  A.  =  12*  15"  16*.5,  Dec.  =  3°  48'  31"  S. 
22*,   3)'s  R.  A.  =  12  17     2  .9,  Dec.  —  4,     2  39  S. 
Castor's  R.A.  =    7  24  24  .4,  Dec.  =32  14    2N. 

Ans.     44°  18'  W. 

24.  Find  the  distance  of  the  moon  from  the  sun,  1839, 
August  12 d,  Greenwich  time  at  mean  noon. 

©'s  R.  A.  =    9*25™51*.72,     Dec.  =  15°  7'51".5  N. 
D  »s  R.  A.  =  11  42  23  .48,     Dec.  =    0  57  27  .9  N. 

Ans.     36°  33'  14". 


§   104.]         '  LONGITUDE.  339 

Lunar  distances. 

25.  Find  the  distance  of  the  moon   from  the   sun,   1839, 
August  14d,  Greenwich  time  at  mean  noon. 

©'s  R.  A.  =    9h  33™ 24*.57,     Dec.  =  14°  31'  28".2  N. 
3>'s  R.  A.  =  13    8  27  .62,    Dec.  =  10  25  54  .5  S. 

Arts.     58°  50'  38" 


340  SPHERICAL    ASTRONOMY.  [CH.  IX. 

Annual  and  diurnal  aberration. 


CHAPTER   IX. 


ABERRATION. 


105.  The  apparent  position  of  the  stars  is  affected 
by  two  sources  of  optical  deception,  so  that  they  are 
not  in  the  direction  in  which  they  appear  to  be. 

The  first  of  these  sources  is  the  motion  of  the  earth, 
and  the  corresponding  correction  is  called  aberration. 

Aberration,  like  the  earth's  motion,  is  either  annual 
or  diurnal. 

106.  Problem.    To  find  the  aberration  of  a  star. 

Solution.  The  apparent  direction  of  a  star  is  obviously  that 
of  the  telescope,  through  which  the  star  is  seen.  Let  S 
(fig.  47.)  be  the  star,  and  O  the  place  of  the  observer  at  the 
instant  of  observation  :  SO  is  the  true  direction  of  the  star, 
or  the  path  of  the  particle  of  light  which  proceeded  from  the 
star  to  the  observer,  and  it  would  be  the  direction  of  the  tele- 
scope if  he  were  stationary.  But  if  he  is  moving  in  the 
direction  OP,  the  direction  of  the  telescope  OT  must  be 
such,  that  the  end  T  was  at  the  point  R,  in  the  line  OS,  at 
the  same  instant  in  which  the  particle  of  light  was  at  this 
point.  The  length  RT  is,  therefore,  the  distance  gone  by  the 
observer  while  the  light  is  describing  the  line  OR. 

If,  then,  we  put 


<§>  107.]  ABERRATION.  341 

Aberration  in  latitude  and  longitude. 

V=z  the  velocity  of  light, 
v  =  the  earth's  velocity, 
/=  TOP  =  RT09 
3I=z  —  ROT=z  the  aberration  from  the  true  place, 

m  =  apghf  (607) 

we  have, 

V:v=z  OR:  TR  =  sin.  I :  —  dlsin.  1" 

dl=z  —  7w  sin.  J.  (608) 

107.  Problem.  To  find  the  annual  aberration  in 
latitude  and  longitude. 

Solution.  The  earth  is  moving  in  the  plane  of  the  ecliptic 
at  nearly  right  angles  to  the  direction  of  the  sun.  Hence  if 
TP  (fig.  48.)  is  the  ecliptic,  T  the  point  towards  which  the 
earth  is  moving,  S  the  true  star,  S1  the  apparent  star, 

©  ss=  the  sun's  longitude, 

A  —  the  star's  longitude,  8  A  —  the  aberration  in  long. 

L  ss  the  star's  latitude,     d  L  =  the  aberration  in  lat. 

we  have 

ST  =1,  SP  =  L, 

long,  of  T  =  ©  —  90°,  PT  =  ©  —  90°  —  A  =  j$ 

PP>  =  dA=  TP—  TP',  *  L  =  SP'  —  SP 

cos.  !T=  cotan.  /tang.  ^  =  cotan.  ( J+  * -O  tanS*  (Ai  —  * -^)j 

whence  tan-«-^>  i  ta"g-(J+/J)>  (609) 

tang.  Jt  tang,  i  %       ' 

29* 


342  SPHERICAL    ASTRONOMY.  [CH.  IX» 

Aberration  in  latitude  and  longitude. 

and,  by  (287  and  288), 

sin.  S  A  sin.  1 1 


nn.(2j1-w#utf)  sin.(2/+<J2)' 


(610) 


or  omitting  tJ  and  <>  /  in  the  denominators,  and  reducing  by 
means  of  (608), 


sm.  2  /  sin.  I  cos.  I 

sin.  j ,  cos.  ^j 


(611) 


COS.  J 

But  cos  /=  cos.  ^j  cos.  Z,  (^12) 

whence  <M  =  w  sin.  ^,  sec.  L  (613) 

=  —  ?w  cos.  (O  —  ^)  sec.  L. 

We  also  have 

.  sin    L        sin.  (Z  +  J^)  /«i>i\ 

SHI.  T  ==  -. =r  =      .      ,  r     =r#  (614) 

bin./         sin  (/-)-<$/)  v       ' 

whence 

sin.  Z  sin.  (/+  <5  /)  tea  sin.  I  sin.  (Z  +  a  Z),        (615) 

sin.  Z    cos.   I 
»nd  ^Iz: i — : --  *  I 

cos.  x.  sin.  jc 
=  —  m  tang,  Z  cos.  I 
z±  —  jw  cos.  -^  t  sin.  Z 
=  —  m  sin.  (o  —  j)  sin.  Z.  (616) 

108.  ProTdcm.   To  find  the  annual  aberration  in  dis- 
tance and  direction  from  the  vernal  equinox. 


<§>  109.]  ABERRATION.  343 

Aberration  from  vernal  equinox. 

Solution.  Let  A  (fig.  48.)  be  the  vernal  equinox,  and  let 
M  =  SA,       $Mz=  aberration  of  M, 
N=z  SAT,   3N=z  aberration  of  N. 

Now  we  have 

*  M  =  $  I  cos.  AST  z=z —. — --=—. — <JJ 

sin.  M  sin.  I 

sin.  0  —  cos.  M  cos.  I  /„„^v 

=  —  m i — „ .  (617) 

sin.  M  v       ' 

But 

cos.  I=z  sin.  ©  cos.  M —  cos.  Q  sin.  M  cos.  N,     (618) 
whence  if  we  put 

B  =  —m  sin.  ©  (619) 

C=z—  m  cos.  O,  (620) 

we  have 

<5 M=z  B  sin. M+C cos.  M cos.  N. 

Again ;  the  triangles  ASS1  and  A  TS'  give  by  (243), 

sin.  AST=  S^O'*  =  _«~Ori°-g 

oj  sin.  J  v       ' 

"■4-  -i,  sin.  iV  C  sin.  iV"  ,„c™ 

^ N  =  m  cos.  ©  -r— : — — .  (622) 

^  sin.  M  sin.  M  v       ' 

109.  Problem.     To  find   the   annual  aberration  in 
right  ascension  and  declination. 

Solution.  If  AT  (fig.  48.)  were   the  equator,   we  should 
have 

D=zSP,    R=zAP, 


344  SPHERICAL    ASTRONOMY.  [CH.  IX. 

Aberration  in  right  ascension  and  declination. 

and  if  we  put 

iVj  ==  SAP,  w  =  obliquity  of  ecliptic, 
we  have 

and  the  triangles  ASP,  AS'P1  give 

sin.  D  =2  s  n  if  sin.  Nt  (623) 

sin.  (D  —  d  D)  =  sin.  (M—  8  31)  sin.  {N1  —  a  N)     (624) 
cos. D$D=z sin.  if  cos.  iV1  S i\r+  cos.  if  sin. iV^  <jil  (625) 
—  J3  sin.  M  cos.  ii  sin.  iVj 

—  C(sin.  iVcos.  N1  —  cos.2  M sin.  Nx  cos.  N), 

and  if  we  put 

A  z=  C  cos.  to  (626) 

6'  =  sin.  M  cos.  iJ^T  sin  Nx  sec.  Z>  (627) 

a'=i — (sin.iVcos.iY^ — cos. 2  if  sin.  iV^  cos.iV)sec.Dsec.w,  (628) 

we  have 

cos.  M  ==  cos.  D  cos.  R  (629) 

cotan.  iV\  =  sin.  R  cotan.  Z>  (630) 

sin.  D  cos.iV,         .      ^ 

sin.  M  cos.  iV\  be ■ - *-  c=  sin.  JD  cotan.  iV\   (631 ) 

sin.  iVj  i   v       / 

z=z  sin.  D  sn.  R  cotan.  Z>  z=  sin.  R  cos.  D 

b'  =  sin.  Z>  cos.  JR  cos.  X>  sec.  D  =  sin.  D  cos.  i2         (632) 

a'=z— [sin.(iV—  iVJ-J-sin.2  if  sin.  Nx  cos.  N]  sec.D  sec. « 

=  [sin.  w  —  sin.2  if  sin.2  Nx  sin.  w]  sec.  D  sec.  w 

—  sin.2  M  sin.  iV^  cos.  Nt  cos.  w  sec  JP  sec.  w 


<§>   109.]  ABERRATION.  345 

Aberration  in  right  ascension  and  declination. 


=  (1 — sin.2  D)  sin.  w  sec.  D  sec.  w  —  sin.  Ms'm.D  cos.  Nt  sec.Z) 
=i  cos.  Z>  tan.  w  —  sin.  JR  sin.  D  (633) 

<>Z>  z=^La'  +  Bb'.  (634) 

Again,  we  have 

cos.  M  z=  cos.  JR  cos.  D  (635) 

cos.  (itf  +  S  M)  =  cos.  (R-j-dR)  cos.  (Z>  +  <*  D) 
cos.  I>  sin.  R$R  —  sin.  Jf  a  ilf —  cos.  Rsm.DSD 
z=  £  (sin.2  M—b>  cos.  J?  sin.  Z>) 
-f-  ^4  (sin.  Mcos.  Mcos.  iVsec.  ^  —  a'  cos.  JR  sin.  D),     (636) 
and  if  we  put 

a=z(s'm.Mcos.  M cos.  Nsec.  w — a!  cos.R  sin.  D)  secD  cosec.  J? 
6=  (sin.2  M —  b'  cos.  R  sin.  X>)  sec.  X>  cosec.  R9 
we  have 

a  cos.  D  sin.  12  —  sin.  M  cos.  if/cos.  iV2  +  sin*  -K  cos-  -B  sin.22> 
-J-  (sin.  if  cos.  ifcT  sin.  N±  —  cos.  R  sin.  D  cos.  Z>)  tan.  w 
zzz  sin.  jR  cos.  R  (cos.2  Z>  -f-  sin.2  D) 
-{-(sin.THsin.iV^cos.jR  cos.D — cos.jR  sin.  M sin. Nx  cos.D)tan.  <*> 
±=  sin.  12  cos.  jR 
a  zs  cos.  J?  sec.  D       •-  (637) 

6  cos.  D  sin.  JR  z=  1  —  cos.2  M —  sin.2  D  cos.2  R 

=  1 —  cos.2  2>  cos.2 12  —  sin.2  D  cos.2  i2 

=  1  —  cos.2  JRizzsin.2  R 

b  =z  sin.  R  sec.  Z>  (638) 

*R=Aa  +  Bb,  (639) 

and  formulas   (619,  620,  632,  633,  634,  637,  638,  639)  agree 


346  SPHERICAL    ASTRONOMY.  [CH.  IX. 

Aberration  in  right  ascension  and  declination. 


with  those  given  in  the  Nautical  Almanac  for  finding  the  an- 
nual aberration. 

110.  Corollary  The  value  of  m,  which  is  used  in  the  Nau- 
tical Almanac,  is 

m  =n  20".3600, 

which  gives 

m  cos.  co  —  20".3600  cos.  23°  27'  36".98  =  18".6768. 

111.  Scholium.  In  the  values  of  the  aberration  in  right 
ascension  and  declination,  each  term  consists  of  two  factors, 
one  of  which  is  the  same  each  instant  for  all  the  stars,  and  the 
other  is  the  same  for  each  star,  during  several  years. 

i 

112.  Corollary.  If  in  (634)  and  (639)  we  put 

i  =z  A  tan.  «  (640) 

B  =z7i  cos.  H  (641) 

-4=  A  sin.  H;  (642) 

they  become 

<3Z?  =  I  cos.  D  —  7i  sin.  Hsin.  R  sin.  D-\-U  cos.  H  cos.  R  sin.  D 
=  i  cos.  D-\-h  cos.  (H+  R)  sin.Z)  (643) 

tR  =  h  sin.  H  cos.  J?  sec.  D  -\-h  cos.  //sin.  J2  sec.  D 

=  A  sin.  (H+  R)  sec.  2>,  (644) 

which  agree  with  the  formulas  in  the  Nautical  Almanac. 

113.  We  have  from  (619-639) 

d 2?=zsec.  D  [ —  m  cos.  w  cos.  ©  cos.  R  —  m  sin.  Q  sin.  R]  (645) 


§  113.]  ABERRATION.  347 

~~Tabie~XLII. 


ss  sec.  D  [ —  J  m  ( cos.  to  + 1 )  ( cos.  ©  cos.  R  -{-  sin.  ©  sin.  R ) 
-f-  J  m  ( I —  cos.  w)  (cos.  O  cos.R  —  sin. ©  sin.  R)] 
:=zsec.Z>[ — m  cos.2  J  w  cos.(jR— ©  )-\-m  sin. 2£to  cos.  (i2-}"0 )]  > 
and  if  we  put 

Q  —  R—  o,     Q'  =  i2+0  (646) 

n  =  —  m  cos.2  J  w,  rc'  -=  7ft  sin.2  J  a>,  (647) 

(645)  beeomes 

S  R  —  sec.  D  (n  cos.  Q  +  ri  cos.  Q'),  (648) 

and  the  values  of  n  cos.  Q  and  ?*'  cos.  Q'  may  be  put  in  tables 
like  Parts  I  and  II  of  Table  XLII  of  the  Navigator. 

Again,  we  have 
$  D=z  sin.  D  (m  cos.  w  sin,  R  cos.  ©  —  m  cos.  R  sin.  0 ) 

—  m  sin.  oi  cos.  ©  cos.  D 

z=  sin.  Z>  [J  m  (cos.  w-j-  l)sin.  Q  —  \m  (1  —  cos.  w)  sin.  Q'] 

—  J  m  sin.  (o  [cos.  (©  +  D)  +  (cos.  ©  —  !>)] 
=sin.Z>[-7wcos.2^cos.)Q+90°)+Jmsin.2Jcucos.(Q/+900)] 

—  \m  sin.  w  [cos.  (0  +  2?)  -f-  cos.  (©  —  D)] 
==  sin.  2>  [—?i  cos.  ( Q  +  90°)  +  »'  cos.  ( Q'  —  90°)] 

-Jm  sin.  co  [cos.  (©+!>)  +  cos.  ( 0  —  !>)],    (649) 
and  the  values  of 
— Jmsin.  wcos.  (©  +  -#)  and  — Am  sin.  to  cos.  (@ —  D) 

may  be  put  in  a  table  like  Part  III  of  Table  XLII.  The 
rules  for  rinding  the  variations  in  right  ascension  and  declina- 
tion are  then  the  same  as  in  the  explanation  of  this  table. 


348  SPHERICAL    ASTRONOMY.  [CH.  IX. 

Table  XLI.~ 

114.  In  constructing  Table  XLII,  the  values  of  m  and  w 
were  taken 

m  =  20",     co  —  23°  27'  28",  (650) 

whence 

n  =  —  19".173,     n>  ==  0".827,  (651) 

—  J  m  sin.  co  —  —  3".9814.  -  (652) 

115.  By  putting 

©  -  J  =  P,  (653) 

we  have,  by  (613  and  615),- 

SL  —  —  m  cos.  (P  —  90°)  sin.  L  (654) 

d  J  =i  —  m  cos.  P  sec.  Z»,  (655) 

so  that  if  the  values  of 

—  m  cos.  P 

are  inserted  in  tables  like  Table  XLI  of  the  Navigator,  the 
variations  of  latitude  and  longitude  are  found  by  the  rule  given 
in  the  explanation  of  this  table. 

116.  If  the  star  is  nearly  in  the  ecliptic,  the  aberration  in 
latitude  may  be  neglected,  and  the  aberration  in  longitude 
will  be  by  (655) 

dJz=z  —  m  cos.  P.  (656) 

117.  Problem.  To  find  the  diurnal  aberration  in 
right  ascension  and  declination. 

Solution.     Let 

v'  ==  the  velocity  of  a  point  of  the  equator,  arising 
from  the  earth's  rotation, 

»'  =  ir^-iir-  (657) 

V  sin.  1" 


§   119.]  ABERRATION.  349 

Aberration  from  the  motion  of  the  star. 

The  velocity  of  the  observer  is  evidently  in  proportion  to  the 
circumference  which  he  describes  in  a  day,  that  is,  to  the 
radius  of  this  circumference,  or  to  the  cosine  of  the  latitude. 

The  velocity  of  the  observer  z=z  v'  cos.  lat. 

Now,  the  diurnal  motion  is  parallel  to  the  equator,  whence 
the  formulas  (613)  and  (616)  may  be  referred  at  once  to  the 
present  case  by  putting 

Z  s=  the  right  ascension  of  the  zenith, 

and  changing  m  into  m!  cos.  lat.,  ©  —  A  into  Z —  R,  and  L 
into  D ;  whence  the  diurnal  aberrations  in  right  ascension 
and  declination  are 

JR  =  —  m'  cos.  (Z—R)  sec.  D  cos.  lat.  (658) 

ti  D  —  —  m!  sin.  (Z  —  R)  sin.  D  cos.  lat.  (659) 

118.  The  value  of  m1  is  nearly 

*»>==<k3L  (660) 

119.  Problem.  To  find  the  aberration  which  arises 
from  the  motion  of  a  planet. 

Solution.  The  most  important  planets  revolve  about  the  sun 
almost  uniformly  in  circles,  and  in  the  plane  of  the  ecliptic. 
At  the  instant,  then,  of  the  light's  reaching  the  earth,  the 
planet  has  advanced  in  its  orbit  by  a  distance  proportioned 
to  its  velocity,  and  to  the  time  which  the  light  takes  in  reaching 
the  earth.  Let  then  S  (fig.  49.)  be  the  sun,  and  OxO[  per- 
pendicular to  O XS  the  path  of  the  planet;  and  put 

vx  =  the  velocity  of  the  planet, 

r  =  OS,     r,  =  0,-S, 
30 


350  SPHERICAL    ASTRONOMY.  [CH.  IX. 

Table  XXXIX. 
we  have 
il^-01001'  =  _%^^i=-Wlcos.Pl.(661) 

But  it  will  be  shown  in  Theoretical  Astronomy  that 
v2  :  v2  —  rx  :  r; 
hence  m2  :  m\  zz  v2  :  v2  zz  rx  :  r 

m  :  m1  —  \Zr±  :  \/r 

mj  z:  m  \/ —  (662) 


\Jzzz-m  V—  cos.  Pj  ;  (663) 


and  this  aberration  being  combined  with  (656)  gives  the 
whole  aberration  in  longitude,  from  which  a  table,  like  Table 
XXXIX  of  the  Navigator,  may  be  constructed. 

120.    Examples. 

1.  Find  the  values  of  log.  A,  log.  B,  h,  II,  and  i  for  May  1, 
1839,  when  ©  zz  40°  23'  52". 

Arts.     log.  A  zz  1.1498B 

log.  B  —  1.1248  R 

h=    19".42 

H  =  226°  40' 

izz  —  6".  13 

2.  Find  the  values  of  log.  a,  log.  6,  log.  a',  log.  6'  for  Al- 
tair  in  the  year  1839. 


§   120.]  ABERRATION.  351 

Annual  aberration. 

Solution. 

R  =  19"  42™  55*         cos.  9.63760  sin.  9.95466  „ 

D  =z  8°  26'  52"  sec.  0.00474  sec.  0.00474 


log.  a  =  9.64234       log.  b  =.  9.95940  n 
R         cos.  9.63760  sin.  9.95166„ 

D         sin.  9.16704  sin.  9.16704         cos.  9.99526 


log.  b'  =  8.80464,  0.13234    9.12170„    w  tan.  9.63747 


0.42927  9.63273 


a'  =  0.56161  log.  a'  =  9.74947 

3.  Find  the  values  of  log.  a,  log.  6,  log.  a7,  log.  b'  for  Regu- 
lus  in  the  year  1839 ;  for  this  star 

R  —  9h  59m  48s,     D  —  12°  45'  7". 

^Lns.     log.  a  =  9.94816n 

log.  b  =  9.71048 

log.  a'  —  9.49516 

log.  &/=9.28122n 

4.  Find  the  numbers  of  the  different  parts  of  Table  XLII 
for  the  argument  7*  20°  —  230°. 

Ans.     12//.32  for  Part  I, 
—  0".53  for  Part  II, 
2".56  for  Part  III. 


352  SPHERICAL   ASTRONOMY.  [CH.   IX. 

Annual  aberration. 

5.  Find  the  number  of  Table  XLI  for  V  20°. 

Ans.     12.9. 


6.  Find  the  aberration  in  right  ascension  and  declination  of 
Altair  for  May  1,  1839. 


Solution.     I. 

A 

1.1498, 

1.1498, 

a 

9.6423 

—  7".93 

a1 

9.7494 

—  6".20 

0.7921n 

0.8992, 

B 

1.1248, 

1.1248, 

b 

9.9594, 

3D  = 

—  0".85 

V 

8.8046 

12".  14 

1.0842 

9.9294, 

33=5".9;=(h39 

:— .$".1% 

II. 

H  +  a  =  162°  44'  +  360°  sin.  9.4725  cos.  9.9800, 

A=19".42  1.2882  1.2882 

2>=8°27'  sec.  0.0047  sin.  9.1670 


a R  =  5".83  a  0S.39  0.7654,     —  2 ".72        0.4352, 

i  cos.  D  =z  —  6 '.06 

t  D  =  —  8",78 


§    120.]  ABERRATION.  353 


Annual  aberration. 


III. 

R  —  ©  =     255°  40'  =  8s  15°  40'        P.  I  =  4".75 
R  +  ©  =  76°+  360  =  2s  16°  +  12*  P.  II  =  0".20 


4'.95     0.6946 
D  sec.     0.0047 


$R  —  5"  =  0s.  33  0.6993 

8*  15°  40'+  3*z=z  11s  15°  40'    P.  I.  —  18".57 


16°+3s=z5*16° 

P.  II.  — 

D 

—  2".85 

—  2".66 

—  3  .38 

0" 

.80 

I*1 

.37 

sin. 

1.2871,, 
9.1670 

@  +  Z>  =  48°  =  ls18° 
©  — D=32°=l'   2° 

0.4541n 

$D       ==  —  8".89 

7.  Find  the  aberration  in  right  ascension  and  declination  of 
Regulusfor  May  1,  1839. 

Ans.     By  Naut.  Aim.  *R  =z      0*.38 

dD=z—  1".87 

By  the  Navigator  SR  —      0*.38 

*!>==— 1".91 

8.  Find  the  aberration  of  Regulus  in  latitude  and  longitude 
for  May  1,  1839. 

Ans.     *  4  ==  6".5 

dL=0".15 

30* 


354  SPHERICAL    ASTRONOMY.  [CH.    IX. 

Aberration  of  the  planets. 

9.  Find  the  aberration  of  Venus  in  longitude,  when  the  dif- 
ference of  longitude  of  Venus  and  the  sun  is  45°. 

Solution.         r  0.0000  0.0000 

rt        ar.  co.  0.1407  J  (ar.  co.)  0.0703 

P  =  45°  sin.  9.8495  20"    log.  1.3010 

Pt  ■=  sin.  9.9902  cos.  9.3214 


0.6927 
—  5"  when  P  x  is  acute,  +  &"  when  Px  is  obtuse, 
—14"  from  Table  XLI      —14" 


lAzn  — 19"  when  P x  is  ac,  = —  9"  when  Px  is  obtuse. 

10.  Find  the  aberration  of  each  of  the  planets  in  longitude, 
when  the  difference  of  longitude  of  the  sun  and  planet  is  15°. 
The  value  of  log.  r±  for  each  of  the  planets  is 

For  Mercury       9.5878  is  the  mean  value, 
Venus  9.8593 

The  Earth  0.0000 
Mars  0.1829 

Jupiter         0.7161 
Saturn  0.9795 

Uranus  1.2829 

Ans.     For  Mercury     — 43"  when  Px  is  acute, 
4"  when  P  x  is  obtuse, 


Venus 

—  41"  when  P \  is  acute, 

3"  when  P  x  is  obtuse, 

Mars 

35" 

Jupiter 

28" 

Saturn 

26" 

Uranus 

24" 

§  120.]  ABERRATION.  355 

Diurnal  aberration. 

11.  Find  the  diurnal  aberration  of  right  ascension  and 
declination  of  Polaris  for  Jan.  1,  1839,  and  latitude  45°,  when 
the  hour  angle  is  0h  30w. 


Solution.          0".31 

9.4914 

9.4914 

45° 

cos. 

9.8495 

9.8495 

D  =  88°  27' 

sec. 

1.5678 

sin. 

9.9998 

0*30"* 

cos. 

9.9963 

sin. 

9.1157 

*'  R  =  —  8".04  =  —  0*.53    0.9050  9  D  ±2  0".03  8.4564 

12.  Find  the  diurnal  aberration  of  ^Ursse  Minoris  in  right 
ascension  and  declination  for  Jan.  1,  1839,  and  latitude  0°, 
when  the  star  is  upon  the  meridian. 

Dec.  of  9  Ursse  Minoris  =  86°  35'. 

Ans.     3R  =  —  0\35 
3'D=z  0, 


356  SPHERICAL    ASTRONOMY.  [CH.  X« 

Refraction  of  a  star. 


CHAPTER   X. 

REFRACTION. 

121.  Light  proceeds  in  exactly  straight  lines;  only  in 
the  void  spaces  of  the  heavens  ;  but  when  it  enters  the 
atmosphere  of  a  planet,  it  is  sensibly  bent  from  its 
original  direction'according  to  known  optical  laws,  and 
its  path  becomes  curved.  This  change  of  direction  is 
called  refraction  ;  and  the  corresponding  change  in  the 
position  of  each  star  is  the  refraction  of  that  star* 

122.  Problem.   To  find  the  refraction  of  a  star. 

Solution.  Let  O  (fig.  50.)  be  the  earth's  centre,  A  the  posi- 
tion of  the  observer,  AEF  the  section  of  the  surface  formed 
by  a  vertical  plane  passing  through  the  star.  It  is  then  a  law 
of  optics,  that 

Astronomical  Refraction  takes  place  in  vertical  planes, 
so  as  to  increase  the  altitude  of  each  star  without  affect- 
ing its  azimuth. 

Let,  now,  ZBH  be  the  section  of  the  upper  surface  of  the 
upper  atmosphere  formed  by  the  vertical  plane,  SB  the  direc- 
tion of  the  ray  of  light  which  comes  to  the  eye  of  the  observer. 
This  ray  begins  to  be  bent  at  B,  and  describes  the  curve  BA, 
which  is  such,  that  the  direction  AC'is  that  at  which  it  enters 
the  eye.     Let,  now, 


$  122.]  REFRACTION.  357 

Ratio  of  sines  in  the  law  of  refraction. 
tp  —  ZAC  z=  the  ^fc's  apparent  zenith  distance, 
r  =i  the  refraction, 

==  the  difF.  of  directions  of  AC  and  BS, 

=  SBL  —  S'CL 
n  =  COZ, 

and  we  have 

LCS'  =  (p  —  ui 
SBL  =  (p  —  u  -f-  r. 

Again,  it  is  a  law  of  optics  that  the  ratio  of  the  sines 
of  the  two  angles  LBS  and  ZAS'  is  constant  for  all 
heights,  and  dependent  upon  the  refractive  power  of  the 
air  at  the  observer* 


Denote  this  ratio  by  w,  and  we  have 
sin.  ((p  —  u  -f-  r) 


sin.  (p 
and  if 


=  n,  (664) 


U  and  R  =z  the  values  of  u  and  r  at  the  horizon, 

we  have 

sinJ^-u +  rl  =     =  cQs> 

sin.  y  \  /         \       / 

whence 

sin.g)  —  sin. (9  —  u-\-r)        1  —  cos.  (U  —  R) 


sin.y  -f-sin.  (<? — w-f-r)        1  +  cos.  (E/"  —  i?) 
tang.  J(m  —  r) 


(666) 


tang.  [9—  J(«— r)] 


=  tang.2J(?7_«)  =  iV,     (667) 


358  SPHERICAL    ASTRONOMY.  [CH.  X. 

Approximate  refraction. 

and  since  \  (u  —  r)  is  small, 

I  (ti  —  r)  =  N  tang.  [<p  —  \  (u  —  r)].  (668) 

Again,  to  find  w,  the  triangle  CO  A  gives 
sin.  (y  —  u)  OA 


sin.  y 


OC 


(669) 


Now  the  point  C  is  at  different  heights  for  different  zenith 
distances  of  the  star ;  but  this  difference  in  the  values  of  OC 
is  small,  and  may  be  neglected  in  this  approximation ;  so 
that 

sin.  (y—  u)  OA  n 

i =  cos.  U  =  —=,  (670) 

sin.  <p  OK 


sin.  (p  —  sin.(<j> — u)  1 — cos.  U 

sin.  (p  -{-  sin.  (y  —  u)        1  -\-  cos.  U 


(671) 


tan.  Jw  =  tang.2  £17  tan.  (y  —  J  w).  (672) 

and  since  u  is  small, 

£  u  =  tang.2  \  U tang,  (<p  —  £  a)  (673) 

which,    compared   with  this   rough   value  of  J  (u  —  r)  from 
(668), 

£  (u  —  r)  =  N  tan.  {$  —  £  u)  (674) 

gives 

*      «  =  i=N^Yu  =  N'r'        '    <675> 

and  if  we  put 

2iV 


(676) 


"~  N'  —  l 
p  =  J(^'-l),  (677) 


$   124.J  REFRACTION.  359 

Table  XII. 

we  have,  by  (668), 

J  (u  —  r)  =  p  r  (678) 

r  =.  m  tan.  (9  —  p  r),  (679) 

and  the  values  of  m  and  p  must  be  determined  by  observation ; 
their  mean  values,  as  found  by  Bradley,  and  adopted  in  the 
Navigator,  are 

m  —  57".035,    p  =  3,  (680) 

by  which  Table  XII  is  calculated. 

123.  The  variation  in  the  values  of  m  and  p  for  different 
altitudes  of  the  star  can  only  be  determined  from  a  knowledge 
of  the  curve  which  the  ray  of  light  describes.  But  this  curve 
depends  upon  the  law  of  the  refractive  power  of  the  air  at 
different  heights ;  and  this  law  is  not  known,  so  that  the 
variations  of  m  and  p  must  be  determined  by  observation. 
At  altitudes  greater  than  12  degrees,  the  mean  values  of  m 
and  p  are  found  to  be  nearly  constant,  and  observations  at 
lower  altitudes  are  rarely  to  be  used. 

124.  The  mean  values  of  m  and  p,  which  are  given  in 
(680),  correspond  to 

the  height  of  the  barometer  nd  29.6  inches,         (681) 
the  thermometer  z=z  50°  Farenheit.     (682) 

Now  the  refraction  is  proportional  to  the  density  of  the  air ; 
but,  at  the  same  temperature,  the  density  of  the  air  is  propor- 
tional to  its  elastic  power,  that  is,  to  the  height  of  the  barom- 
emeter.     If  then 


360  SPHERICAL    ASTRONOMY.  [CH.  X. 

Table  XXXVI. 

h  =  the  height  of  the  barometer  in  inches, 
r  —  the  refraction  of  Table  XLI, 
$  r  =  the  correction  for  the  barometer ; 

we  have 

r  :  r  +  dr  =  29.6  :  h  (683) 

29.6  9  r  =  (h  —  29.6)  r  (684) 

(A— 29.6)  .- 

*r=      29.6      r>  (685) 

whence   the   corresponding   correction  of  Table  XXXVI  is 
calculated, 

Again,  the  density  of  the  air,  for  the  same  elastic  force,  in- 
creases by  one  four  hundredth  part  for  every  depression  of  1° 
of  Fahrenheit;  hence  the  refraction  increases  at  the  same 
rate,  so  that  if 

<5;  r  =  the  correction  for  the  thermometer, 

f  -s  the  temperature  in  degrees  of  Fahrenheit, 

we  have 

whence  the  corresponding  correction  of  Table  XXXVI  is  cal- 
culated. 

125.    Examples. 

1.  Find  the  refraction,  when  the  altitude  of  the  star  is  14°, 
and  the  corrections  for  this  altitude,  when  the  barometer  is 
31.32  inches,  and  the  thermometer  72°  Fahrenheit. 


§   126.]  REFRACTION.  361 

Corrections  for  barometer  and  thermometer. 


Solution.  15 '.035  log.  1  75614 

76°  tan.  0.60323 


1  st  app.  r  =z  228".7  =2  3'  48".7   2.35937 

15".035    1.75614 

76°—  3  r  =  75°  48'  34 "    tan.  0.5971 1 


2d  app.  r  —  226"  ==  3'  46"     2.35325  2.353 

31.32  —  29.6  *=  1.72      0.235  50— 72=— 22  1.342„ 

29.6      ar.co.  8.529   400  ar.  co.  7.398 


6r  zzzlS"  1.117   a'r=  — 12"  1.093* 

2.  Find  the  refraction,  when  the  altitude  of  the  star  is  50°, 
and  the  corrections  for  this  altitude,  when  the  barometer  is 
31.66  inches,  and  the  thermometer  36°. 

Ans.     The  refraction  =z  48" 

Correction  for  barometer  r=     3" 

Correction  for  thermom.   =*     2" 

3.  Find  the  refraction,  when  the  altitude  of  the  star  is  10°, 
and  the  corrections  for  this  altitude,  when  the  barometer  is 
27.80  inches,  and  the  thermometer  32°. 

Ans.     The  refraction  —  5'  15" 

Correction  for  barometer  = —  19" 

Correction  for  thermom.  =       15" 

126.  Problem.    To  find   the  radius  of  curvature  of 
the  path  of  the  ray  of  light  in  the  earth's  atmosphere. 
31 


362  SPHERICAL    ASTRONOMY.  [CH.  X. 

Path  of  the  ray  of  light. 

Solution.  By  the  radius  of  curvature  is  meant  the  radius  of 
the  circular  arc,  which  most  nearly  coincides  with  the  curve. 
Now  this  radius  may  be  found  with  sufficient  accuracy,  by  re- 
garding the  whole  curve  AB  as  the  arc  of  a  circle  ;  and  if  we 
put 

rt  =  the  radius  of  curvature, 
Rx  z=  OA  =  the  earth's  radius, 
we  have 

AC  :  Rx  =  sin.  u  :  sin.  (^  +  u),  (687) 

or,  nearly, 

AB  :  R1  z=  u  sin.  1"  :  sin.  y 

AB=R*uaia-1".  (688) 

sin.  y 

Again,  the  radii  of  the  arc  AB,  which  are  drawn  to  the 
points  A  and  B,  are  perpendicular  to  the  tangents  AS'  and 
BS,  so  that  the  angle  which  they  make  with  each  other  is 

S'AS  =  r  5 
that  is,  r  is  the  angle  at  the  centre,  which  is  measured  by  the 
arc  AB,  consequently 

AB  =  rx  sin.  r  z=  1\  r  sin.  l/y,  (689) 

whence 

(690) 

(691) 
(692) 

(693) 


r 

r  sin.  q> 

But,  by  (678), 

u 

=  7r, 

whence 

ri 

_  7JR, 

sin.  (p 

so  that  at  the  horizon 

as  in  (225,  226). 

ri 

=  7  Rx 

$    127.]  REFRACTION.  363 

Dip  of  the  horizon. 

127.  Problem.     To  find  the  dip  of  the  horizon. 

Solution.  The  dip  of  the  horizon  is  the  error  of  supposing 
the  apparent  horizon  to  be  only  90°  from  the  zenith,  whereas 
it  is  more  than  90°.  If  O  (fig.  51.)  is  the  centre  of  the  earth, 
B  the  position  of  the  observer  at  the  height  AB  above  the 
surface,  O'  the  centre  of  curvature  of  the  visual  ray  BT, 
which  just  touches  the  earth's  surface  at  T,  BH1  perpendic- 
ular to  O'Bj  is  the  direction  of  the  apparent  horizon  and 

a  H  —  HBH1  =  OBO>  —  the  dip. 

The  triangle  BOO'  gives 

BO1 :  OO1  =  sin.  BOO' :  sin.  §  H=z  sin.  BOH' :  sin  *  H, 

or,  siRce         BO'  —  1  BO  nearly,  and  OO'  =z6BO, 

and  d  H  and  BOH'  are  small, 

7:6=  BOHf    dH 

AH' 
3H=%  BOH'  =  f  -         ,         ,  (694) 

7  7   AO  sin.  I"  v       ' 

But,  by  (227),  we  have,  if  we  put 
R  =  AO,  h=z  AB 

f  AH'  =  f-  V(  J  R  h) 

=  2  V(f  R  h  (695) 

whence 

and 

log.  a  //  =  log.  2  —  log.  (a/%R)  —  log.  sin,  1"  +  £  log.  A 

=  1.77128  +  J  log.  7i,  (697) 

which  is  the  same  with  the  formula,  given  in  the  preface  to 
the  Navigator,  for  calculating  Table  XIII. 


364  SPHERICAL    ASTRONOMY.  [CH.  X. 

Dip  of  the  sea. 


128.  Problem.   To  find  the  dip  of  the  sea  at  different 
distances  from  the  observer. 

Solution.  Let  O  (fig.  52.)  be  the  centre  of  the  earth,  B  the 
observer  at  the  height 

h  —  AB  (in  feet) 

above  the  sea,  and  A1  the  point  of  the  sea  which  is  observed 
at  the  distance 

d  —  A  A1  (in  sea  miles)  =  AOA1 

from  B ;  and  let 

M  ==  the  length  of  a  sea  mile  in  feet. 
If  the  radius  OA  is  produced  to  B'9  so  that 
A  B1  ==  AB, 

the  point  B1  will  be  elevated  by  refraction  nearly  as  much  as 
the  point  A1.  But  the  visual  ray  BB'  will,  from  the  equal 
heights  of  B  and!?',  be  perpendicular  to  the  radius  OC,  which 
is  half  way  between  B  and  B',  so  that  the  dip  of  B'  is,  by  (694), 

6 B  ==  |  BOG  =2-AOA'z=fd.  (698) 

The  dip  of  the  point  A!  will  be  greater  than  B'  by  the  angle 

i  —  B'BA, 

which  it  subtends  at  B,  and  which  is  found  with  sufficient  ac- 
curacy by  the  formula 

A I  Til  7» 

=  TO  =■<"*■•*'  (699) 


AB  —  Md 


M  sin.  T  d 
But,  by  (228), 


(700) 


^=w  <™> 


§     131.]  REFRACTION.  365 

Twilight. 

1        =J0800_ 

M sin.  T       n  R  sin.  1'  ?  v       1 

so  that  the  dip  of  A1  is 

9  A  —  f  d  +  0.56514  £  (703) 

- 
which  is  the  same  with  the  formula,  given  in   the  preface  to 

the  Navigator,  for  calculating  Table  XVI. 

129.  Refraction,  by  elevating  the  stars  in  the  horizon, 
will  affect  the  times  of  their  rising  and  setting ;  and 
the  star  will  not  set  until  its  zenith  distance  is 

90°  +  horizontal  refraction, 

and  the  corresponding  hour  angle  is  easily  found  by 
solving  the  triangle  PZB  (fig.  35.) 

130.  Another  astronomical  phenomenon,  connected 
with  the  atmosphere,  and  dependent  upon  the  combi- 
nation of  reflection  and  refraction  is  the  twilight^  or 
the  light,  before  and  after  sunset,  which  arises  from  the 
illuminated  atmosphere  in  the  horizon.  This  light  be- 
gins and  ends  when  the  sun  is  about  18°  below  the 
horizon  ;  so  that  the  time  of  its  beginning  or  ending  is 
easily  calculated,  from  the  triangle  PZB  (fig.  35.) 


131.  Examples. 

1.  Find  the  dip  of  the  horizon,  when  the  height  of  the  eye 
is  20  feet. 

Ans.     264"  =  4'  24". 
31* 


366  SPHERICAL    ASTRONOMY.  [CH.  X. 

Dip. 

2.  Find  the  dip  of  the  sea  at  the  distance  of  3  miles,  when 
the  height  of  the  eye  is  30  feet. 

Solution.  fK8  =  f=s  17.3 

0.56514  X  \°         =  5'.6 


dip      ==  7'. 

3.  Find  the  dip  of  the  sea  at  the  distance  of  2J  miles,  when 
the  height  of  the  eye  is  40  feet. 

Ans.     10'. 

4.  Find  the  dip  of  the  sea  at  the  distance  of  J  of  a  mile, 
when  the  height  of  the  eye  is  30  feet. 

Ans.     68'. 


§  133.]  PARALLAX.  367 

Parallax  in  altitude. 


CHAPTER    XL 


PARALLAX. 


132.  The  fixed  stars  are  at  such  immense  distances 
from  the  earth,  that  their  apparent  positions  are  the 
same  for  all  observers.  But  this  is  not  the  case  with  the 
sun,  moon,  and  planets ;  so  that,  in  order  to  compare 
together  observations  taken  in  different  places,  they 
must  be  reduced  to  some  one  point  of  observation. 
The  point  of  observation  which  has  been  adopted  for 
this  purpose  is  the  earth's  centre  ;  and  the  difference 
between  the  apparent  positions  of  a  heavenly  body,  as 
seen  from  the  surface  or  the  centre  of  the  earth,  is 
called  its  parallax. 

133.  Problem.   To  find  the  parallax  of  a  star. 

Solution.  Let  O  (fig.  53.)  be  the  earth's  centre,  A  the 
observer,  £the  star,  and  OSA,  being  the  difference  of  direc- 
tions of  the  visual  rays  drawn  to  the  observer  and  the  earth's 
centre,  is  the  parallax.  Now  since  SAZ  is  the  apparent  ze- 
nith distance  of  the  star,  and  SOZ  is  its  distance  from  the 
same  zenith  to  an  observer  at  O,  the  parallax 

OSA=p 

is  the  excess  of  the  apparent  zenith  distance  above  the  true 
zenith  distance.     If,  then, 


368  SPHERICAL   ASTRONOMY.  [CH.  XI. 

Parallax  in  altitude. 

%  =z  SAZ,  R  =  OA  z=z  the  earth's  radius, 
r  =  OS  r=  the  distance  of  the  star  from  the  earth's  centre, 
we  have  P  :  R  =n  sin.  Z :  sin.  p, 

R  sin.  z 


or 

sin.  p  =. , 

r 

(704) 

or 

R  sin.  z 
p  zz: . 

r  sin.  1" 

(705) 

134. 

Corollary,  If  P  is  the  horizontal 

parallax,  we  have 

sin.  P  —  — , 
r 

(706) 

or 

p-    R  - 

r  sin.  \"  ; 

(707) 

whence 

sin.  p  =  sin.  P.  sin.  z9 

(708) 

or 

p  =          P.  sin.  z. 

(709) 

which  agrees  with  (564)  and  Tables  X.  A.,  XIV,  and  XXIX, 
are  computed  by  this  formula,  combined,  in  the  last  table,  with 
the  refraction  of  Table  XII. 

135.  Corollary.  In  common  cases,  the  value  of  the 
horizontal  parallax  can  be  taken  from  the  Nautical 
Almanac ;  but,  in  eclipses  and  occultations,  regard  must 
be  had  to  the  length  of  the  earth's  radius,  which  is 
different  for  different  places.  The  curvature  of  the 
earth  is  such  that  the  radius  diminishes  as  we  recede 
from  the  equator  proportionally  to  the  square  of  the  sine 
of  the  latitude ;  the  whole  diminution  at  the  pole  being 
about  ^tj<jth  part  of  this  radius. 


§  136.]  PARALLAX.  369 

Reduction  of  parallax. 

Now  the  horizontal  parallax  is,  by  (707),  proportional  to 
the  earth's  radius,  so  that  it  diminishes  at  the  same  rate,  from 
the  equatorial  value  which  is  given  in  the  Nautical  Almanac. 
Hence,  if 

$  R  ==  the  diminution  of  R  for  the  latitude  L, 

dP  —  that  of  P, 

R  s=  the  radius  at  the  equator, 

P  =  the  parallax  at  the  equator, 

we  have  <3  R  —  -^  R  sin.2  L 

=  ^22(1  —  cos.2Z)  (710) 

aP  =  TrforPsin.2JE, 

=  1faP(\  —  cos.2Z),  (712) 

and  if  P  is  expressed  in  minutes,  while  $  P  is  expressed  in 
seconds,  (711)  becomes 

*  P  in  seconds  =  T^  (P  in  minutes)  (1  —  cos.  2  i),  (712) 

which  agrees  wifh  the  formulas  for  calculating  the  reduction 
of  parallax  given  in  the  explanation  to  Table  XXXVIII  of 
the  Navigator. 

136.  In  reducing  delicate  observations  to  the  centre  of  the 
earth,  it  must  be  observed  that  the  centre  is  not  exactly  in  the 
direction  of  the  vertical.  Thus,  if  O  (fig.  54.)  is  the  earth's 
centre,  P  the  pole,  A  the  observer,  Z  the  zenith,  ZAL  the 
vertical,  Z1  the  point  where  the  radius  OA  produced  meets 
the  celestial  sphere,  Z'  is  called  the  true  zenith,  and  Z  the 
apparent  zenith.  The  angle  ZAZ1,  which  is  the  difference 
between  the  polar  distance  of  the  true  and  apparent  zenith 
is  called  the  reduction  of  the  latitude,  and  must  be  subtracted 
from  the  angle  ALE,  or  the  latitude  to  attain  the  angle  AOE, 


370  SPHERICAL,    ASTRONOMY.  [CH.  XI. 

Reduction  of  the  latitude. 

or  the  direction  of  the  observer  from  the  earth's  centre.  The 
angle  AOE  is  called  the  reduced  latitude,  and  is  to  be  substi- 
tuted for  the  latitude  in  reducing  delicate  observations  to  the 
centre  of  the  earth. 


'  137.  Problem.    To  find  the  redaction  of  the  latitude. 

Solution.    Draw  OB  (fig.  54.)  parallel  to  AE;  with  OA 
as  a  radius  describe  the  arc  AR.     The  angle 

§L  =  AOB=z  OAL 

is  the  reduction  of  the  latitude,  and  is  so  small,  that  the  arcs 
AB  and  AR  may  be  regarded  as  straight  lines,  and  the  tri- 
angle ABR  as  a  right  triangle ;  and  since  the  sides  AB  and 
AR  are  perpendicular,  respectively,  to  AL  and  AOf  we 
have 

SL  =  ABR. 

If  now,  we  put 

m  z=z  the  difference  of  the  polar  and  equatorial  radii 

divided  by  the  equatorial  radius  =z  -g-^,         (712) 

we  have,  by  neglecting  the  very  small  terms  m^L2,  <5jL3,  m3, 

R=:OA  =  R(l  —  m  sin2  L)  (713) 

OB  =  R  [1  —  m  sin.2  (L  +  d  L)] 

—  R[l  —  m  (sin.2  L  -f  sin.  9L  cos.  L)2] 

=z  R(l — msin.2.L — 2msin.JLcos..Lsin.<2.L)  (714) 

RAB  —  R—OBz=z2mR  sin.  L  cos.  L*L  (715) 

AB  =  R>  sin  <>  L  =  R  sin.  9 L  (1  —  m  sin.2  L)     (716) 

,r         .^  BR        2ms'm.  L  cos.  L       ._„. 

tMng.»L  =  mn.tL=-[5=    l_m,m,r,     <™> 


$   139.]  PARALLAX.  371 

Reduction  of  the  latitude. 

whence 

sin.  a  X  —  m  sin.  a  X  sin.2  X  =  2  m  sin.  X  cos.  X      (718) 
sin.  a  X  z=  2  m  sin.  X  (cos.  X  -f-  i  sm»  *  -£'  sin.  X) 
=  2  m  sin.  X  (cos.  X  -f-  sin.  J  a  X  sin.  X) 

=  2  m  sin.  X  cos.  (X,  —  J  a  X)  (719) 

ii 

a  X  =  ■*: — --  sin.  X  cos.  (X  —  J  a  X) 
sm.l"  v  2        ' 

==   „     2   t,    sin.  X  cos.  (X  —  J  a X),  (720) 

5  sin.  I7  ' 

from  which  <?X  is  easily  calculated. 

138.  Corollary.    By  putting  in   the  last  term  of  (719),  for 
sin.  ^X,  the  approximate  value 

sin.  $  L  =z2m  sin.  X  cos.  X,  (721) 

it  becomes 

sin.  a  X  —  2  m  sin.  X  (cos.  X  +  m  sin.2  X  cos.  X) 

=  2  w  sin.  X  cos.  X  (1  +  m  sin.2  X)  (722) 

a  X  =  f  cosec.  1'  sin.  X  cos.  X  ( 1  +  ■$$  sin.2  X)    (723) 

139.  Corollary.    We  have,  by  (56), 
1-J-tan.Xtan.ax  _         l-\-ta.n.LtaLn.SL 


cot.(X  —  aX)  = 


tan.X  —  tan. ^X  ~     tan.X(l — cot.Xtan.ax) 

T  1  +  tanff.  X  tang,  a  X  ,^,x 

=  cotan.  X  r-S ^— 2-jt-»  (724) 

1  —  cotan.  X  tang,  ax  v       ' 


372  SPHERICAL    ASTRONOMY.  [CH.  XI. 

Reduction  of  the  latitude. 

and  if  we  put 

n  =  2  m  (1  +  m  sin.2  L)  z=2m-\-2m2  sin,2  Z 
—  2m-\-m2  —  m2  cos.  2  Z,  (725) 

we  have. 

tang.  Z  =  sin.  9  Z  =  n  sin.  Z  cos.  Z  (726) 

1  +  tang.  Z  tang.  9  L  =z  1  +  n  sin.2  Z  (727) 

1  —  cotan.  Z  tang,  d  L  =  1  —  ft  cos.2  Z  (728) 

cotan.  {L  —  *L)  =  cotan.  Z  (1  +  nsin.2  X.)  (1  —  w  cos.2  Z)-1 

=  cotan.  Z  (I  +  rc  sin.2  Z)  (1  +  ncos.2  Z+ £n2  cos.4  Z) 
=  cot.Z[l+n(sin.2Z+cos.2Z)+w2cos.2Z(sin.2Z+cos.2Z)] 
zz: cotan.  Z  (1  -f-w  +  in2  cos.2  Z) 
z=  cotan.  Z  (1  +  ft  +  £  n2  +  J  n2  cos.2  Z) 
:zzcotan.Z(l+2w  +  3wi2  +  m2  cos.2  Z),  (729) 

and  the  term  m2  cos.2  Z  is  so  small,  that  it  may  be  neglected, 
whence  we  have 

cotan.  (Z  —  *  Z)  =  ( 1  +  2  m  +  3  m2)  cotan.  Z 

=  1.0067  cotan.  Z,  (730) 

which  agrees  with  the  formula  given  in  the  explanation  of 
Table  XXXVIII  in  the  Navigator,  and  which  must  be  calcu- 
lated by  means  of  Tables  of  7  places  of  decimals. 

140.  Problem.  To  find  the  parallax  in  latitude  and 
longitude. 

Solution.  Let  Z  (fig.  55.)  be  the  zenith,  P  the  pole  of  the 
ecliptic,  and  M1  the  apparent  place  of  the  place  of  the  body, 
whose  parallax  is  sought,  and  M  its  true  place.     Let  also 


§  140.]  PARALLAX.  373 

Parallax  in  latitude  and  longitude. 

N  z=  PZ  =.  the  zenith  distance  of  the  pole, 

=  the  altitude  of  the  nonagesimal, 

A  =a  90°  —  ZM'  —  the  apparent  altitude, 

L  ==  90°  —  PM  =  the  true  latitude  of  the  body, 

D  =  ZPM  =  the  true  diff.  of  long,  of  the  body 

and  the  zenith, 

P  z=z  the  horizontal  parallax, 

p  z=  P  cos.  A  =z  MM'  z=z  the  parallax  in  altitude, 

3D  =  ZPM  —  ZPM  =z  the  parallax  in  longitude, 

3  L  —  PM1  —  PM  =  the  parallax  in  latitude, 

L'  =z  L  +  dL. 

The  triangles  PMM>  and  ZPM'  give 

_        »  sin.  M        p  sin.  I?  sin.  (D  4-dD) 

d  D  z=  £ —  =  x- -p- 2 £ '- 

cos.  Z,  cos.  A  cos.  JLi 

—  P  sin.  B  sec.  £  sin.  (D  +  9  D)  (731) 

Again,  the  triangles  ZPM  and  ZPM'  give 

__      sin. Zi — cos.  JBsin. (;!+»)      sin.Z/ — cos.Bsm.A  /m,ne%. 

cos.  Zzz  — : — —v- — L£i  = : — - (732) 

sin.  B  cos.  (A  -f-p)  sin.  B  cos.  -A 

whence 

sin.L  cos.  A — sin.Z/cos.  (A-\-p)  zzz  cos. JB sin.  (A-\-p)  cos.  A 

— cos.  B  sin.  A  cos.(A-{-p) 
=  cos.  B  sin.  p.  (733) 

But      sin.Z*  =:sin.(Z/ — s  L)zzz\s'm.L'cosJL — cos.L'sm.dL 
izzsin.Z' — cos.Z/sin.  <*  L — 2  sin.Z/sin.2  J  <?  Z 
nzsin.^— cos.i/sin.^jL— Jsin.Z'sin.s^Z  (734) 
cos.(^i  -\-p)  =  cos.il — sin.  A  sin.p — £  cos.  ^L  sin.2p,        (735) 
32 


374  SPHERICAL    ASTRONOMY.  [CH.  XI. 

Parallax  in  latitude  and  longitude. 

whence 

cos.  JL'  cos.  A  *  L  z=  —  (cos.  B — sin. A  s'\n.L')p 

-\-%sin.L'cos.A  (psm.p  —  dZsin.^Zi).  (736) 
But 

sin.4r=cos.Psin.Z/-|-sin.Pcos  Z/cos.  (D  +  ^D)  (737) 
whence 

cos.  B  —  sin.  A  sin.  L!  z=  cos.  B  —  cos.  B  sin.2  U 

—sin..B  sin.Z'cos.Z'sin. (Z>+dI>) 

zrcos.JBcos.2!,'—  sin.Bsin.Z/cos.Z/cos.(I>+^Z>),(738) 

and  also       cos.  B  —  sin.,4  sin.X/rzcos.iW 'cos.Z'cos.^l,  (739) 
io  that,  for  a  first  approximation, 

3L  =  —  cos.  M'p  (740) 

^ L  sin.  5  L  =z  cos.2  M'  p  sin.  p  (741 ) 

psin.pnK5.Lsin.  $  L=z  sin.2  M '  p  sin.  p 

—  sin.  M1  cos.  U  sin.  <J  D  .p 

cos.  L'  sin.  B  sin.  D  d  D  i~mc%\ 
P>          (742) 


cos.  A 
and  (736)  becomes 
*L=z  —  cos.JBcos.Z/.P-(-sin.JBsin.JL/.P[cos.(D  +  aI>) 

+  Jsin.  !><>Z)] 

=z  —  cos.  B  cos.Z/  .P+sin.Psin.Z/cos.(Z>-f  J  d  ^J^,  (743) 

and  formulas  (731)  and  (743)  agree  with  the  rule  in  the  Navi- 
gator [B.  p.  404]. 

141.  Problem.     To  find  the  parallax  in  right  ascen- 
sion and  declination. 


§   142.]  PARALLAX.  375 

Parallax  in  right  ascension  and  declination. 

Solution.  Formulas  (731)  and  (743)  may  be  applied  imme- 
diately to  this  case,  by  putting 

B  =  the  altitude  of  the  equator  =  the  co-latitude, 
L  ==  the  true  declination, 
U  =  the  apparent  declination, 

D  =  the  right  ascension  of  the    body  diminished  by 
that  of  the  zenith  =  the  hour  angle  of  the  body. 

&L  =  the  parallax  in  declination, 
$D  ==  the  parallax  in  right  ascension. 

142.  Corollary.  The  value  of  3  L  may,  in  this  case,  be 
found  by  a  somewhat  different  process,  which  is  quite  con- 
venient when  the  altitude  of  the  body  is  required  to  be  calcu- 
lated. Draw  PN  to  bisect  the  angle  MPM',  and  draw  MH 
and  M1  H1  perpendicular  to  PN,  and  we  have  nearly 

d  L  ==  HIP  =  HN+H'N' 

=  MN  cos.  N+M'N  cos.  N 

=  {MN  +  M'N)  cos.  N  =  MM'  cos.  N 

=  P  cos.  A  cos.  N.  (744) 

Now,  in  the  triangle  PZN,  we  have 

PZ  =  co-lat.  ZPN  —D  +  i5Dt 

and  may  take 

PN  =  90°  —  L,  ZN  =  90°  —  A, 

so  that  PZ,  ZPN,  and  PN  are  given,  to  find  ZN  and  N. 
This  method  of  calculating  the  parallax  in  right  ascension  and 
declination  is  precisely  that  used  in  [B.  p.  443]  for  calculating 
from  the  relative  parallax  the  corrections  for  right  ascension 
and  declination. 


376  SPHERICAL    ASTRONOMY.  [CH.  XI. 

Apparent  diameter. 

143.  The  apparent  diameter  of  a  heavenly  body  is 
the  angle  which  its  disc  subtends. 

144.  Problem.  To  find  the  apparent  semidiameter 
of  a  heavenly  body. 

Solution.  Let  O1  (fig.  56.)  be  the  centre  of  the  heavenly 
body,  A  the  observer,  and  A  T  the  tangent  to  the  disc  of  the 
body.     The  angle  TAO1  is  the  apparent  semidiameter.     Let 

Rx  ==  OT 

°    =  OAT 

r    =z  AO', 


we  have 

sin.  a 

OT 

'  AO' 

r 

(745) 

Hence, 
body, 

by  (fig. 

53.), 

if  A  is 

the 

apparent 

altitude 

of  the 

> 

sin.  a 

R, 

sin 

■P 
~'P) 

(746) 

R  cos 

.(A 

a 

R1 

sec, 

.(A- 

-P)- 

(747) 

145.  Corollary.  If  2  is  the  horizontal  semidiameter,  we  have 

which  is  also  the  semidiameter,  as  seen  from  the  earth's  cen- 
tre. 

Now  Rx  —  0.2725  R  (749) 

R    =      3.67/2,  (750) 

whence    log.  -^  =  9.43537,  (ar.  co.)  =  0.5646,  (751) 


§  147.]  PARALLAX.  377 

Augmentation  of  semidiameter. 

so  that  formula  (748)  agrees  with   [B.  p.  443.    No.  10  of  the 
Rule]. 

146.  Corollary.  If  <J  a  is  the  augmentation  of  the  semi- 
diameter  for  the  altitude  A,  we  have  by  (747),  and  putting 
A1  =  A  —  b, 

PRi  _  PRlcos.(A'—p) 

~  .Rcos. (^1-|~P)  ~  12cos.il' 

s   cos.  (A1  —  p) 
cos.  A' 

sin.  p  sin.  A' 

=  2  +  2 *-— 

1  cos.  A' 

=z  2  +  2  sin  p  sin.  A  (752) 

<J  a  —  2  sin.  P  sin.  A  =  2  P  sin.  1"  sin.  A 

=  L?l  p2  gin.  1"  sin.  4.  (753) 

Now  for  the  mean  horizontal  parallax  of  57'  30",  we  have 
log.  ^  P2  sin.  1"  =  1.19658  (754) 

4r  P2  sin.  1"  =5  15.72,  (755) 

It 

agreeing  very  nearly  with  the  explanation  to  Table  XV  of  the 
Navigator. 

147.  Corollary.  The  augmentation  can  also  be  calculated 
without  determining  the  altitude.     Thus,  from  (752) 

/COS.     A  ^\  /-y-„v 

32* 


378  SPHERICAL    ASTRONOMY.  [CH.   XI. 

Augmentation  of  semidiameter. 

But  from  (fig.  55.)  and  (731) 

sin.(D+ sD).  cos.  (L—dL)    __ 

cos.  A  =  sin.  ZM'  = - ! — .  '  „ — ^ '    (757) 

sm.  Z  v       ' 

rrnr  Sm«   &  C0S-   -^  /~^v 

cos.  4'  =  sm.  ZM  = ; — = —  (758) 

sin.  Z  v       ' 

cos.  A      1_sin.  (Z>  +  <?Z)).cos.  (Z*  —  <sZ) 
cos.il'  sin.  Z>  cos.  L 

_  cos.  jPcos.  (L—  3L)SD       cos.(L—3L)__ 
cos.  Z, .  sin.  D  cos.  i 

_  P.cos.ff.sin.ffsin.(Z>  +  <?l>)        cos.(Z-<?Z) 
cos.  Zr  sin.  D  \      cos.  X 

Now  the  latitude  of  the  moon  is  so  small,  that,  in  the  first 
term,  we  may  put 

cos.  1=1,  (760) 

which  gives  by  (756),  and  putting 

H=z  ZP.  cos.  D  sin.  B  (761) 

H<=4C0S'{L-dV--l\  (762) 

\       cos.  L  J  ' 

**=  H+  H cos.  DdD  +  H> 

=  H+H.P.cos.Dsm.B  +  H' 

=  H+~+H'.  (763) 

Now  we  have  by  (761)  and  (762) 

H=zi2.p.  [sin.  (B  +  D)  +  sin.  ( B  —  D)]       (764) 

H'=  -2- (tang,  i .^ i  +  cos.  d L  —  1),  (765) 

and  formulas  (763  to  765)  agree  with  the  method  of  calculat- 
ing the   augmentation  of  the   semidiameter  given   in  Table 


<§>  148.]  PARALLAX.  379 

Augmentation  of  semidiameter. 

XLIV  of  the  Navigator.     The  three  first  parts  of  this  table 
are  calculated  for  the  value  of  2, 

2  =  16'  =  960", 

whence  £s.P  =  8".18. 

The  fourth  part  of  the  table  is  the  correction  which  arises 
from  the  difference  between  the  actual  value  of  2  and  that  as- 
sumed in  the  three  former  parts.     If  we  put 

9'  a  —  the  value  of  9  a  for  2  =  16', 

we  have,  by  (755)  and  (748), 

9a  :?'•<,  ='**''.  (167)2  (766) 

22 

d  0  — &  a 

256 


^+(£-1)" 


2  2  _  256 

.  d'  a 


-     "  '        256 

=  ,.+£±!«>£=>S»..  <767) 

as  in  the  explanation  of  this  table. 

148.    Examples. 

1.  Find  a  planet's  parallax  in  altitude,  when  its  horizontal 
parallax  is  25",  and  its  altitude  30°. 

Am.     22". 

2.  Find  the  moon's  parallax  in  latitude  and  longitude,  when 
her  horizontal  parallax  is  59'  10".3 ;  her  latitude  3°  7'  19"  S., 


380  SPHERICAL    ASTRONOMY.  [CH.  XI. 

Parallax  in  latitude  and  longitude. 

her  longitude  44°  36'  16";  the   altitude  of  the  nonagesimal 
37°  56'  14",  its  longitude  25°  27'  16",  the  latitude  of  the  place 

43°17/18//N. 

Solution. 

Reduced  parallax  =  59'  10".3  —  5".3  =  59'  5//=3545" 
Reduced  latitude  =  43°  17'  18"—  IT 27":=  43° 5' 51" 

D  =  44°  36'  16"  —  25°  27'  16"  =  19°  9' 
3545  3.54962  3.54962  3.550 

37°  56'  14"  sin.  9.78873  cos.  9.89691    sin.  9.789 

3°    7'  19"  sec.  0.00064  3°    7'  19"  cos.  9.99936 


* 

3.33899   46' 32" 

3.44589 
9.99899 

19°  9' 

sin.  9.51593  3°  53' 51" 

12' 

2.85492   4630" 

3.44552 
19°  15' 

19°  21' 

3" 

sin.  9.52027  3°  53  41" 

sin.  8.831 

3D  =12' 

2.85926 

— 2' 20" 

log.  9.975 

19°  21' 

3" 

2.145 

*  Z,  =  44'  10" 

3.  Find  the  moon's  parallax  in  latitude  and  longitude,  when 
her  horizontal  parallax  is  60' £".9;  her  latitude  1°  30'  12"  N., 
her  longitude  130°  17',  the  altitude  of  the  nonagesimal  85°  14', 
its  longitude  125°  17',  the  latitude  of  the  place  46°  IF  28".4  N. 

Ans.     Parallax  in  longitude  ==  5'  18" 

Parallax  in  latitude      =s  3'  30",5. 


§   148.]  PARALLAX.  381 

Augmentation  of  semidiameter. 

4.  Calculate  the  parts  of  Table  XLIV,  when  the  argument 
of  the  first  part  is  3s  19°  =  109°  ;  that  of  the  second  12".4, 
the  moon's  true  latitude  1°  20'  N.,  the  moon's  parallax  in  lati- 
tude 50',  the  sum  of  the  three  first  parts  13",  and  the  moon's 
horizontal  semidiameter  14'  50". 

Solution.         8".  1845  sin.  109°  —  7".74  =  Part  I. 

(12"  4>* 

Part  III  =  960"  [sin.  50'  tang.  1°  20'  —  1  +  cos.  50'] 
=  960"  [sin.  50/  tang.  1°  20'  —  2  sin.2  25'] 
=  960"  [0.00023]  en  0".22. 

,  30  50"  X  1'  10"  13"X 30.83X1.17 


Part  IV  =— 13"  X 


256'  ~~  256 

=  — 1".83. 


5.  Calculate  the  parts  of  Table  XLIV,  when  the  argument 
of  the  first  part  is  2*  16°,  that  of  the  second  15."5,  the  moon's 
true  latitude  3°  S.,  the  moon's  parallax  in  latitude  307,  the  sum 
of  the  three  first  parts  11",  and  the  moon's  horizontal  semi- 
diameter 15'  20". 

A?is.     Part  I    =      7".94 

Part  II  =      0  .25 

Part  III=:—0.48 

Part  IV  :=—  0  .90 

6.  Calculate  the  number  of  Table  XV,  when  the  altitude 
is  45°. 

Ans.     11". 


382  SPHERICAL   ASTRONOMY.  [cH.   XI. 

Augmentation  of  semidiameter. 

7.  Calculate  the  augmentation  of  the   moon's  semidiameter 
in  Example  2;  when  the  horizontal  semidiameter  is  16' 50". 

Solution.  Part  I     =  6".87  +  2".58  z=      9".45 

Part  II    =  0  .09 

Part  III  :=  —1  .02 


sum     =?      8".52 
Part  IV  z=  0  .91 


augmentation    =      9".43 

8.  Calculate  the  augmentation  of  the  moon's  semidiameter 
in  Example  3,  when  the  horizontal  semidiameter  is  15'  30". 

Ans.     M?',83. 


<§>  151.]  eclipses.  383 

Solar   eclipse* 


CHAPTER   XII. 

ECLIPSES. 

149.  A  solar  eclipse  is  an  obscuration  of  the  sun, 
arising  from  the  moon's  coming  between  the  sun  and 
the  earth  ;  and  occurs  therefore  at  the  time  of  new 
moon. 

It  is  central  to  an  observer,  when  the  centre  of  the 
moon  passes  over  the  sun's  centre.  It  is  total,  when 
the  moon's  apparent  disc  is  larger  than  the  sun's,  and 
totally  hides  the  sun.  It  is  annular,  when  the  moon's 
apparent  disc  is  smaller  than  the  sun's,  but  is  wholly 
projected  upon  the  sun's  disc. 

The  phase  of  an  eclipse  is  its  state  as  to  magnitude. 

150.  An  occultation  of  a,  star  or  planet  is  an  eclipse 
of  this  star  or  planet  by  the  moon. 

A  transit  of  Venus  or  Mercury  is  an  eclipse  of  the 
sun  by  one  of  these  planets. 

151.  Problem.  To  find  when  a  solar  eclipse  will  take 
place. 

Solution.  Let  O  (fig.  57.)  be  the  sun's  centre,  and  Ol  the 
moon's  centre  at  the  time  of  new  moon,  and  let 

I?  z=  the  latitude  of  the  moon  at  new  moon 
==  OO^ 


384  SPHERICAL    ASTRONOMY.  [CH.  XII. 


When  a  solar  eclipse  will  happen. 


Let  ON  be  the  ecliptic,  and  N  the  moon's  node,  so  that  N01 
is  the  moon's  path.     Let 

N  =  the  inclination  of  the  moon's  orbit  to  the  ecliptic  ; 

Draw  OP  perpendicular  to  the  moon's  orbit,  and  if,  when  the 
moon  arrives  at  P,  the  sun  arrives  at  O,  the  least  distance  of 
the  centres  of  sun  and  moon  is  nearly  equal  to  O'P.  Now 
the  triangle  OPO1  gives 

OP  —  p  cos.  N=p  —  P(l  —  cos.  N) 

=  f*  —  2  p  sin.2  I  N  =  /»  —  i  p  sin.2  N 
n  =  ratio  of  the  sun's  mean  motion  divided  by  the  moon's 

=  TV  nearly,     (768) 
we  have  OO'  =  n  X  OxP  =  nfi  sin.  N. 

Draw  O'R  perpendicular  to  OP,  and  we  have  nearly 
OR=  OP  —  OP  =i  OO'  sin.  N 
=  n  p  sin.2  N. 
Hence 

OP  —  p  _  ( J  +  n)  /» sin.2  iV:=  P-±'ftP  sin.2  iV.    (769) 

The  apparent  distance  of  the  centres  of  the  sun  and  moon 
is  affected  by  parallax,  and  the  true  distance  is  diminished 
as  much  as  possible  for  that  observer,  who  sees  the  sun  and 
moon  in  the  horizon,  and  OP  vertical,  in  which  case  the 
diminution  is  equal  to  the  difference  of  the  horizontal  paral- 
laxes of  the  sun  and  moon.     Let,  then, 

P  z=z  the  moon's  horizontal  parallax, 
w  =  the  sun's  horizontal  parallax, 
4  =.  the  apparent  distance  of  the  centres, 
we  have 
the  least  apparent  dist.  =  OP  —  (P  —  n) 

~p—  &p8in.*N—  P+n.    (770) 


$  152.]  eclipses.  385 

When  a  solar  eclipse  will  happen. 

Now,  an  eclipse  will  take  place,  when  this  least  apparent 
distance  of  the  centres  is  less  than  the  sum  of  the  semidiame- 
ters  of  the  sun  and  moon.     Thus,  let 

s  =z  the  moon's  semidiameter, 

a  s=  the  sun's  semidiameter. 

In  case  of  an  eclipse,  we  must  have 

p  _  £  /s  sin.  2  N  —  P  +  7r<s  +  cr,  (771) 

or  ?<p  —  7r  +  s  +  (7+T2"^  sin-2  N-  (772) 

152.  Corollary.  We  have,  by  observation, 

the  greatest  value  of  P  =      6V  32", 
the  least  value  =      52'  50", 

the  mean  value  ==      57'  11", 

the  greatest  value  of  ^    =  9", 

the  least  value  =  8", 

the  greatest  value  of  s    =      16'  46'', 
the  least  value  =       14'  24", 

the  mean  value  —       15'  35", 

the  greatest  value  of  <*    •=.      16'  18", 
the  least  value  =      15'  45", 

the  mean  value  —       16'    1", 

the  greatest  value  of  N  —  5°  20'    6", 
the  least  value  =z  4°  57'  22", 

the  mean  value  zzz  5°    8;  44". 

Now,  in  the  last  term  of  (772)  we  may  put  for  N  its  mean 
value,  and  for  p  its  mean  value  obtained  by  supposing  it  equal 
to  the  preceding  terms,  which  gives 

33 


386  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Limits  of  a  solar  eclipse. 

p=P  —  7U  +  s  +  o=8&  38"  =  5318"     (773) 
^=3102" 
sin.  N=  sin.  5°  8'  44"  =  0.09,  sin.2  N=  0.008 

T\p  sin.a  iV  =  25;/,  (774) 

whence  (772)  becomes 

p  <  P  —  tt  +  f  +  a  +  25".  (775) 

153.  Corollary.  If,  in  (775),  the  greatest  values  of  P,  5, 
and  a,  and  the  least  value  of  n  are  substituted,  the  limit 

p  <  1°  34'  52" 

is  the  greatest  limit  of  the  moon's  latitude  at  the  time  of  new 
moon,  for  which  an  eclipse  can  occur. 

154.  Corollary.  If,  in  (775),  the  least  values  of  P,  s,  and 
o,  and  the  greatest  value  of  n  are  substituted,  the  limit 

p  <  1°  23'  15" 

is  the  least  limit  of  the  moon's  latitude  at  the  time  of  new 
moon,  for  which  an  eclipse  can  fail  to  occur. 

155.  Problem.  To  find  the  places  where  a  given 
phase  of  a  solar  eclipse  is  first  and  last  seen. 

Solution.  The  distance  of  the  centres  of  the  sun  and  moon 
will  first  be  reduced  to  a  given  apparent  distance  jf  at  that 
place  where  the  moon  is  vertically  above  the  sun  and  the 
lower  limb  of  the  moon  just  beginning  to  rise.     Let 

P  —  the  relative  horizontal  parallax  of  the  sun  and  moon, 

=  P  —  *u,  (776) 

in  which  it  is  advisable  to  take  for  P  its  reduced  value  for  the 


§  155.]  eclipses.  387 

Places  where  solar  eclipse  begins  and  ends. 

latitude  of  45°,  because  the  latitude  of  the  required  place  is 
not  known. 

For  the  time  of  new  moon,  let 

D  =  the  moon's  declination, 

d  ==  the  sun's  declination, 

R  ==.  the  diff.  of  right  ascension  of  sun  and  moon, 

—  the  moon's  right  ascension  —  the  sun's, 

Dx  z=  the  relative  hourly  motion  in  declination, 

ae  the  moon's  motion  —  the  sun's, 

R1  —  the  relative  motion  in  right  ascension. 

Let  S  (fig.  58.)  be  the  sun,  M the  moon,  MM'  the  moon's  relative 
path,  that  is,  the  path  which  it  would  describe  if  the  sun  were 
stationary,  and  the  moon's  motion  were  the  relative  motion  ;  let 
SP  be  perpendicular  to  MM',  and  N  be  the  north  pole.  The 
zenith  of  the  place  is  in  the  line  SMZ,  which  joins  the  cen- 
tres of  the  sun  and  moon,  and  at  a  distance  SZ  of  about  90° 
from  them.     Let 


t  == 

the  angle  CSP 

k  = 

MSP 

p  = 

SP. 

Join  NP  and  draw  PR 

\  perpendicular  to  NS, 
RPC  =  i 

we  have 

PNR 

Rx          PNR 

CR    ~~ 

Dx  ~~  PRtzn.i 

1 

1 

"sin.  NR  tan.  i       cos 

.  D  tan.  t 

388  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Places  where  a  solar  eclipse  is  first  seen. 

whence 

tan.  is        Dl     -  (777) 

R  x  cos.  D  y       ' 

p  =  CS.  cos.  i—  (D  —  d)  cos.  i  (778) 

PJRzzrpsin.  s,  CR  z=:ps'm.  i  tan./,  PC  —  p  tan. i. 

Let  then 

£  zzz  the  interval  between  the  moon's  passing  from  P  to  C, 

CJR        »  sin.  i  .  *±L**i 


-!>,-   i>,    —"    , 

v  ,%/; 

let 

»  sin.  i 
c=           '     X  3600" 

('80) 

£  (in  seconds)  =  c  tan.  i. 

(781) 

Again, 

let       ^'  =  !£#  =z  the  true  dist.  of  centres  of  sun 

and 

moon, 

we  have                        J'  —  j  -\-  P' 

(782) 

cos.  k  S=  — ; 

J' 

(783) 

MP  s=  p  tan.  A;, 

(784) 

let                       t  —  time  of  describing  ilffP, 

we  have              *  zz=                 =  c  tan.  A; 

(785) 

"  a  —  itftfiV  =  —  i  zp  *s 

(786) 

the  positive  values  of  a  being  reckoned  towards  the  east,  so 
that  the  upper  sign  corresponds  to  the  beginning,  the  lower  to 
the  end  of  the  eclipse. 


§  157.]  eclipses.  389 

Places  where  eclipse  begins  and  ends. 

Finally,  if        L  =  the  latitude  =  90°  —  ZN 

h  =  the  hour  angle  after  noon  =  ZNS, 

the  triangle  ZNS  gives,  by  §  39, 

sin.  L  =  cos.  d  cos.  a  (787) 

tan.  a  #.«*•». 

tan.  h  = : — y.  (788) 

sin.  a 

156.  Corollary.    The  value  of  4  is  for  the  beginning  or 
ending  of  an  eclipse, 

4  =  s  +  a  ;  (789) 

for  the  beginning  or  ending  of  total  darkness  in  a  total  eclipse, 

j  =  s  —  0',  (790) 

for  the  forming  or  breaking  up  of  the  ring  in  an  annular 
eclipse, 

J  =  a  —  s  ;  (791) 

for  the  central  phase, 

J  =  0.  (792) 

157.  Problem.  To  find  the  places  for  which  a  given 
phase  of  the  eclipse  is  seen  at  sunrise  or  sunset. 

Solution.  Let  M  (fig.  59.)  be  the  centre  of  the  true  moon, 
at  any  time  after  the  first  formation  of  the  phase  J  and  before 
its  end,  S  that  of  the  sun,  m  that  of  the  apparent  moon  af- 
fected by  relative  parallax.  Since  the  sun  and  moon  are  in 
the  horizon,  we  have 

Mm  tea  P1, 

also  m  S  =.  4. 

33* 


390  SPHERICAL,    ASTRONOMY.  [CH.  XII. 

Places  where  eclipse  begins  at  sunrise  or  sunset. 

The  zenith  Z  is  in  the  line  Mm,  at  the  distance 
ZS  —  90° 

from  S,  let  N  be  the  north  pole.  Join  MN,  and  draw  Ng 
perpendicular  to  NS,  and  the  right  triangle  NMG  gives,  by 
putting 

D0  z=  the  declination  of  g 
cos  R  =  tang.  Ng  cotan.  MN 

*r  tang.  D  cotan.  2>0  =  ^^,  (793) 

whence,  by  (287)  and  (52), 

sin.(Z>0  —  D)        1  — cos.  22  ^     ,  ^ 

.     ,     °     n  (  3=  T-j ^  xac  tang.  2  J  R,    (794) 

sm.(Z?-f  Z>0)        1  +  cos.  22  6     2      '    v       ' 

or,  since  D0  —  D  and  R  are  small, 

D0—  D  =i±;R2  sin.  1"  sin.  2  D  (795) 

D0  =  D  +  |  jR2  sin.  1"  sin.  2  Z>.  (796) 

Let  now,  z0  zr  g-tf,     y0  =  i^^,  S=MSg, 

and  we  have  z0  z=  D0  —  d  (797) 

yQ  —  R.cos.  D0  (798) 

tan.  S  =  ^  (799) 

4Mb  a:0  sec.  A  (800) 

Now  since    Zilf  and   ZS  are  nearly  quadrants,  they  are 
nearly  parallel  at  their  extremities  If  and  S,  so  that  if 

b  z=z  MSZ  =  m  MS,  (801) 

P'-l-j  P'  —  J 

and  q  =  _^       ql  _  __^9  (g02) 


<§>  161.]  ECLIPSES.  391 

Places  where  a  phase  is  seen  at  sunrise  or  sunset. 

we  have  sin.  J b  =  V (g~^2/^'~g,),  (803) 

whence  the  triangle  NZS  gives 

ZNS  —S^m  (804) 

sin.  JLr=cos.  (S^fm)  cos.  d,  tan.  hz=  — tan.(&=pm)  cosec  d.  (805) 

158.  Corollary.  Since  M  m  may  be  taken  on  either  side  of 
MS,  there  are  two  places  for  each  place  My  except  when 

J  =  J  -f  P't  (806) 

which  corresponds  to  the  beginning  or  to  the  ending  of  the 
phase,  or  when 

J  <P'  —  J.  (807) 

159.  Corollary.  When  the  nearest  approach  4'  of  the  true 
centres  is  less  than  P' —  ^,  the  places  at  which  the  phase  J 
are  seen  in  the  horizon  are  upon  two  different  oval  curves. 

160.  Corollary.  When  the  nearest  approach  4'  of  the  true 
centres  is  greater  than  P' —  j,  the  places  at  which  the  phase  4 
are  seen  in  the  horizon  are  all  upon  one  curve,  which  intersects 
itself,  and  is  formed  like  a  figure  8  much  distorted ;  and  in 
this  case  this  curve  is  the  northern  or  the  southern  boundary 
of  the  eclipse. 

161.  Corollary.  The  values  of  xQ  and  y0  might  be  found 
more  easily,  but  less  accurately,  by  the  formulas 

tang,  k  =  -  (808) 

*  =  p  sec.  k  (809) 


392  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Limits  of  a  solar  eclipse  on  the  earth. 

Sz=:  —  i^Jc  (810) 

x  =  J'  cos.  8,  y0  ==  4'  sin.  S,       (811) 

and  from  the  approximate  value  of  L,  obtained  by  this  pro- 
cess, an  accurate  value  of  P'  may  be  found,  which  may  be 
used  in  the  calculation  by  the  process  before  given. 

162.  Problem.  To  find  the  curves  of  extreme  north- 
ern and  southern  places  at  which  a  phase  is  seen. 

Solution.  When  the  nearest  approach  is  greater  than  P'  —  ^, 
one  of  the  limiting  curves  is,  as  in  §  130,  the  northern  or 
southern  portion  of  the  rising  and  setting  curve,  accordingly 
as  the  moon  passes  to  the  north  or  to  the  south  of  the  sun. 
The  other  limiting  curve  consists  of  those  places,  at  which 
the  nearest  approach  of  the  apparent  centres  is  equal  to  4\ 
and  these  are  the  places  which  compose  both  the  limiting 
curves,  when  the  nearest  approach  is  less  than  P'  —  ^.  The 
eastern  and  western  limiting  curves  are  always  those  of  rising 
and  setting,  and  at  the  points  where  the  rising  and  setting 
curves  cease  to  be  the  limiting  curves,  the  phase  J  is  one  of 
nearest  approach,  and  at  the  same  time  is  in  the  horizon.  We 
have,  then,  only  to  consider  at  present  the  places  where  the 
phase  4  is  one  of  nearest  approach. 

For  this  purpose,  let  M  (fig.  60.)  be  the  true  moon's  centre, 
m  the  apparent  relative  moon's  centre,  S  the  sun's  centre, 
N  the  north  pole,  Z  the  zenith  of  the  place  ;  draw  m  r,  Mg 
perpendicular  to  N&.     Let 

D1  z=z  the  declination  of  m 

R'  zs  m  NS 

%  =  Zm,  M=z  NmZ,  h  =  ZNS 


§  162.]  eclipses.  393 


Limits  of  a  solar  eclipse  on  the  earth. 


MNm  =  relative  parallax  in  right  asc.  =  R  —  R1 

—  P>  cos.  L  sec.  D  sin.  (h  —  R)         (812) 

x=zrnhz=z  D — D'—  relative  paral.  in  dec.  —  P'  sin.  z  cos.  M=. 

c±  P1  [sin.  L  cos.  D  —  cos.  £  sin.  Z>' cos.  (h—  R  )]  (813) 

y z=h M=  (R  —  R)  cos. D  =  P'  cos.  L  sin.  (h  —  R') 

—  P>  sin.  |  sin.  M  (814) 

Now,  by  the  diurnal  motion,  the  angle  h  will  increase  for 
the  instant  <U,  of  an  hour,  by  the  quantity 

15°  U, 

and  the  changes  in  the  other  terms  of  x  and  y  will  be  too 
small  to  be  sensible  in  these  small  quantities  ;  so  that  the  in- 
crements of  x  and  y.will  be,  by  (13)  and  (15), 

dx  =  15°  P'  #t  cos.  L  sin.  D1  sin.  (h— R)~15°  y  st  sin./)'  (815) 
d  y  =  15°  P>  H  cos.  L  cos.  (h  —  R)  (816) 

—  15°  P'  d  t  (cos.  z  cos.  D1  —  sin.  %  sin.  D1  cos.  M ) 

m  15°  P  d  t  cos.  s  cos.  D— 15°  2 :  Jt  sin.  Z>'.  (817) 
Again,  if 

u=Sr9  v  —  mr,  i/=rSM}  (818) 

we  have 

u  =  ^  cos.  i'  =  x  —  z0,  v  —  J  sin.  #  =  y0  —  y,     (819) 

and  if  w  m'  is  the  apparent  relative  orbit  of  the  moon,  it  must 
be  perpendicular  to  Sm,  because  m  is  the  point  of  nearest 
approach.  Hence  if  m!  is  the  place  of  the  moon  at  the  end  of 
the  instant  S  ty  we  have 

$u  z=l  Sx  —  dx0  =z  —  mm'  sin.  i' 

=  15°dtsin.D'  (y0  —  js'in.i')  —  9x0        (820) 


394  SPHERICAL   ASTRONOMY.  [CH.  XII. 


Limits  of  a  phase  upon  the  earth. 


d  v  3=  <$yo  —  dyzzzmm1  cos.  i1  as  — l5°P'd 1  cos.  z  cos.  D1 

+15°  (x0  +  D  cos.  i>)*t  sin.  D'  +  ty0,  (821) 
whence         m  m'  sin.  %'  —  —  d  u  cos.  i1  =  dv  sin.  zv 
z=  — 15°  dt  sin.  D'  (y0  cos.  fl — ^sin.  i'  cos.  i')  +<5a:0  cos.  i'  (822) 
=      15°a^sin.Z>/(a:0  sin.  I'-J-^sin.i'  cos.  2V) 

+  dy0  sin.  i1  —  15°  P'  $  t  sin.  i'  cos.  z  cos.  D1. 

Now  Z)'  differs  so  little  from  d,  that  d  may  be  substituted 
for  it  in  this  equation,  and  we  have  also 

— 9.  —  the  hourly  motion  in  relative  declination  =  D1 

d  * 

♦  y 

— -  sec.  D  =  the  hourly  motion  in  relative  dec.  =  Rlt 

d  t 

and  if  we  put 

__  R.cos.D  D, 

ATW^V'    B-T^^T'  (823) 

(822)  becomes,  by  dividing  by  15°  $t  sin.  1", 

P'  sin.  It  cos.  z  cos.  cZ  =  (A  -f-  z0  sin.  d)  sin.  t' 

_  ( b  —  #0  sin.  d)  cos.  t'.     (824) 
Let  now  ;.  and  v  be  so  taken,  that 

A  -\-  x0   sin.  cf  =  i  P1  cos.  z  cos.  d  cos.  * 
P  —  yQ  sin.  d  =.*  P1  cos.  z  cos.  e?  sin.  *,         (826) 
and  (824)  becomes 

cos.  z  sin.  i'  —  2  sin.  {%'  —  *)  (827) 

Asin.(ev  —  *)  .  4   ..      /000\ 

cos.  z  == r-^- ; -  =  a  cos.  r  —  1  sin.  y  cot.  1'.     (S28) 

sin.  ^/ 

To  find  t',  its  value  may  be,  at  first,  assumed  as  equal  to  i,  as 


§  162.]  eclipses.  395 

Limits  of  a  phase  upon  the  earth. 

it  is  nearly,  because  the  true  relative  orbit  PM  is  nearly  parallel 
to  the  apparent  relative  orbit  m  m'.  Hence  u  and  v  are  found 
by  (819),  and  thence 

D1  =  d  qp  u  (829) 

R1   =  ±  v  sec.  D1  (830) 


i>0  —  D  +  i(R  —  JR)2sin.2D 

(831) 

y    —  (R  —  R)  cos.  D0 

(832) 

x    =D0-D> 

(833) 

V 

tan.  M—  - 

X 
X 

(834) 

/QO£r\ 

sin.  z  —                _  _ 
P'  cos.  M 

[poo  J 

and  from  this  value  of  z,  i'  may  be  found  by  means  of  (828), 
which  gives 

cos    Z 

cot.  i  —  cot.*— — r — ,  (836) 

x  sin.  *  x       ' 

or  if  9  is  taken,  so  that 

.  cos.  z  ;_ 

sin.  ^  sa  Vol (837 

2  *  cos.  v  v       ' 


2  x  cos.  r  sin.2  <p 
sin.  v 


cot.  #  =rf  cot.  v — r- — — *-  (838) 


=;  cot.*  (1  —2  sin.2  <p)  =  cot.  v  cos.  2  (p,    (839) 

whence  new  values  of  z  and  il^  may  be  computed.  Then  the 
triangle  NMZ  can  be  solved  by  the  usual  process,  and  will 
give  the  values  of 

h  —  R'  =  ZNM  and  L  =  90°  —  NZ. 


396  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Limits  of  a  phase  upon  the  earth. 

163.  Corollary.    There  are  two  points,  m  and  m',  at  which 
we  should  have 

j  =  S  m  =  Sm'} 

and  therefore  two  zeniths,  Z  and  Z{ r,  which  correspond  to  the 
two  values  (829-835). 

164.  Corollary.    If  t  is  the  time  of  the  phase   J  counted 
from  the  middle  of  the  eclipse, 

p  =  SP 

the  perpendicular  upon  the  orbit,  and 

k  =  MSP,  k'  =  Mm  S, 
we  have,  nearly,  by  (785), 

MP       ptzn.k  p  t  ,__ 

tan.  kf  =  -=-  =  *  =     /  840 

Jf  =  <—  t)3F#,  (842) 

which  gives  a  rough  method  of  computing  Z  and  M. 

165.  Problem.     To  find  the  duration  of  a  phase  upon 
the  earth. 

Solution.  At  the  first  and  last  points  we  have 

Z=90°,     PM—P', 

cos.  k'  =  F^t  (843) 

t  z=  semiduration  =  c  tan.  k  (844) 

c.PM       c.P'   .      _  /oj^x 

=  TT  =  Hp"  Sm#  (845) 


§  168.]  eclipses.  397 

Central  eclipse. 

166.  Problem,     To  find  the  places  where  the  eclipse 
is  central. 

Solution.  For  these  places  m  and  £  coincide,  so  that 

MS  —  4  —  P'sin.  Z  (846) 

sin.^=  ^,  (847) 

so  that  in  the  triangle  ZSN>  the  two  sides  ZS,  NS,  and  the 
included  angle  >S'  are  given  to  find  NZ  and  ZNS, 

167.  Corollary.  For  one  place  the  eclipse  will  be  central  at 
noon,  and  for  this  place  we  have,  obviously, 

J  z=z  diff.  dec. 

sm.Z=±  (848) 

L  =  d  +  Z  (849) 

west  long,  of  place  =  app.  Greenw.  time  of  cent,  eclipse. 

168.  Problem.  To  calculate  the  time  of  the  beginning 
or  ending  of  a  given  phase  of  a  solar  eclipse  for  a  given 
place. 

Solution.  Find  for  a  supposed  time  near  the  required  time, 
such  as  the  time  of  new  moon,  the  relative  parallaxes  in  right 
ascension  and  declination  of  the  sun  and  moon,  and  their  rela- 
tive right  ascension  and  declination.  Hence  their  apparent 
relative  right  ascension  and  declination  is  found  by  simple 
addition  or  subtraction. 

34  . 


398  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Time  of  a  solar  eclipse  for  a  given  place. 

Let         D  z=l  their  apparent  relative  declination, 

R  zzzz  their  apparent  relative  right  ascension, 
d  r=  the  sun's  declination, 

W  —  the  distance  apart  of  their  apparent  centres, 
j  =  the  phase, 
A  zzzz  the  angle  which  W  makes  with  the  parallel  of 


so  that 


If 

the  supposed  time  is  that  of  the  beginning  or  ending  ©f  the 
phase.  But  if  W  differ  from  J,  find  another  apparent  distance 
W'  of  the  centres  for  a  time  a  little  after  the  former  one. 
Then  we  have  W —  W  :  W  —  A  z=  diff.  of  supposed 
times  :  the  correction  which  is  to  be  added  to  the  first  sup- 
posed time  to  obtain  the  required  time.  If  this  correction  is 
large,  a  new  computation  must  be  made,  using  the  time  just 
obtained  as  a  new  supposed  time. 

169.   Corollary i    The  time  of  the  phase   of  an   eclipse  or 
occultation  might  also  be  calculated  by  the  following  process. 

Let         jR  L  zzz  the  relative  hourly  motion  in  right  ascension, 
D  x  zzzz  that  in  declination  ; 

then  let  S  (fig.  62.)  be  the  centre  of  the  sufij  and  M  that  of 
the  moon  at  the  supposed  time,  CS  the  hour  circle,  A  the 


declination,  and  we  have 

D 

=z  W  sin.  A,  R  cos.  d  —  W  cos.  A              (850) 

R  cos.  d                                         Jr^„  v 
tan,  A  = — — —                                        (851) 

W  =±  D  cosec.  AzzziR  cos.  d  sec.  A.   (852) 

W    z=  J, 

§  170.]  eclipses.  399 

Time  of  a  solar  eclipse  for  a  given  place. 

moon's  centre  at  the   beginning  of  the  phase,  B  at  the  end  ; 
we  have,  then, 

^o,™*-  ™       CM      R  cos.  d  ,0,*™ 

tan.  CSM  =  tan.  S=  -—= - —  (853) 

Co  U 

CT  D 

tan.  i  —  tan.  CMI  —  tan.  FSI—  ~  =  -f> ~3       (854) 

CM       R1co$.d      K       ' 

S31  =  W—y  cosec.  S=x  sec.  #  (855) 

SP  =  p  =  Wcos.  PSM=z  Wcos.  (S+  i)    (856) 

cos.  k\—  cos.  P,SU  =  cos.  PSB  z=z  £  (857) 

a  =z  ASM—  S  +  i  +  k  (858) 

b  =2  BSM  =k  —  (S+  i).  (859) 

Then  let      tx  =1  the  interval  of  moon's  passing  from  A  to  M, 

t2  =  the  time  from  M  to  Bt 
and  we  have 

4 If  cos.  i         PPsin.  ^o^itf  cos,  t 

W  sin.  a  cos.  i 


y  x  cos.  A; 

W  sin.  b  cos.  £ 

2/  x  cos.  /c 


(860) 
(861) 


170.  Corollary.  This  method  maybe  used  by  substituting 
latitude  and  longitude  for  declination  and  right  ascension, 
and  in  this  case  the  sun's  latitude  is  zero,  so  that  the  formulas 
agree  with  the  rule  in  the  Navigator  [B.  p.  425]. 


400  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Magnitude  of  an  eclipse. 

171.  Corollary,  For  the  beginning  or  end  of  the  eclipse  the 
phase  is 

J  —  the  sum  of  the  horizontal  semidiameters  of  the 
sun  and  moon  increased  by  the  augmentation 
of  the  moon's  semidiameter.  (862) 

For  the  beginning  or  end  of  total  darkness  in  a  total  eclipse, 

J  =  difF.  of  semidiam.  -f-  aug.  of  J)  's  semidiam.     (863) 

For  the  formation  or  breaking  up  of  the  ring  in  an  annular 
eclipse, 

J  =z  difF.  of  semidiam.  —  aug.  of  D's  semidiam.     (864) 

172.  Problem.    To  find   the  greatest  magnitude  of 
the  eclipse  at  any  place. 

Solution      Let 

D  sz  the  relative  apparent  declination  at  the  beginning 
of  the  phase  Jf 

A  —  the  angle  which  the  line  joining  the  centre  of  the 
apparent  sun  and  moon  makes  with  the  circle  of 
declination  at  this  time, 

D'  and  A'  ==  the  values  of  D  and  A  at  the  end  of  the 
phase  J, 

j0  z=z  the  nearest  approach  of  the  centres, 

we  have,  by  (fig.  61.),  in  which  MNM1  is  the  moon's  apparent 
relative  orbit,  S  the  sun,  SD  the  circle  of  declinations, 

SM z=z  SM>  zzz  Jy     SN  =  j0, 

D  D' 

sin.  A  —  — ,  sin.  A'  =  — ,  (865) 

J0  =  J  cos.  %  (A'  —  A).  (866) 


$  174.]  ECLIPSES.  401 

Lunar  eclipse. 

173.  The  calculation  of  occupations  is  the  same  as 
that  of  solar  eclipses,  except  that  the  star  has  no  parallax, 
and  its  disc  is  insensible.  The  calculation  of  transits  of 
planets  over  the  disc  of  the  sun  is  the  same  as  that  of  a 
solar  eclipse,  except  that  the  planet  is  to  be  substituted 
for  the  moon. 

174.  Problem.  To  find  when  a  lunar  eclipse  will 
happen. 

Solution.  The  solution  is  the  same  as  in  §  168,  except  that 
the  semidiameter  of  the  earth's  shadow  at  the  distance  of  the 
moon  is  to  be  substituted  for  that  of  the  sun  ;  and  the  change  in 
the  position  and  apparent  magnitude  of  the  moon  from  parallax 
may  be  neglected,  because  when  the  earth's  shadow  falls  upon 
the  moon,  the  moon  is  eclipsed  to  all  who  can  see  it.  Now  if 
S  (fig.  61.)  is  the  sun,  E  the  earth,  GF  the  semidiameter  of 
the  sun's  shadow  at  the  moon,  we  have 

the  app.  semi.  ==  FEG  ==  EFL  —  EIF  ~  P  -—  EIF 

=  P  —  (KES  —  EKI) 

=  P  —  G  +  *j 

or  rather,  this  would  be  the  apparent  semidiameter,  if  it  were 
not  for  the  earth's  atmosphere,  which  increases  the  breadth  of 
the  shadow  about  g^th  part ;  so  that 

the  app.  semidiam.  =  f^-  (P  —  a  -f-  tf), 

and  therefore,  in  order  that  an  eclipse  must  happen,  we  must 
have,  by  (762), 

P  =  the  latitude  at  the  time  of  full  moon, 

P<iHP  +  "  —  a) +  *  +  &(*  s^2  AS  (867) 

34* 


402  SPHERICAL    ASTRONOMY.  [CH.   XII. 

Lunar  eclipse. 

175.  Corollary.  In  the  last  term  of  (867),  we  may  put  for  N 
its  mean  value,  and  for  p  its  mean  value  obtained  by  supposing 
it  equal  to  the  preceding  terms,  which  gives 

f  S6  57'  35"  z=  3455",     ^  p  =  2015" 
sin.2  N  =  0.008,     TV  p  sin.2  N  ==  16", 
whence  (867)  becomes 

/»<**  (**  +  *  —  *)  +  *  +  16"-  (868) 

176.  Corollary.  If,  in  (868),  the  greatest  values  of  P,  n, 
and  5  are  substituted,  and  the  least  value  of  af  the  limit 

£<  63'  45" 
is  the  greatest  limit  of  the  moon's  latitude  at  the  time  of  full 
moon,  for  which  an  eclipse  can  occur. 

177.  Corollary.  If,  in  (868),  the  least  values  of  P,  n,  and 
s  are  substituted,  and  the  greatest  value  of  of  the  limit 

/J  <  51'  57" 

is  the  least  limit  at  which  an  eclipse  can  fail  to  occur. 

178.  Problem.  To  calculate  when  a  given  phase  of 
a  lunar  eclipse  will  occur. 

Solution.     Let 

D  z=z  the  relative  declination  of  the  moon  referred  to 

the  centre  of  the  shadow  at  time  of  full  moon, 
d  =i  the  declination  of  the  centre  of  the  shadow  —  — 

sun's  declination, 
R  =  the  relative  right  ascension, 

z=  D'sR.A.-Q's  R.  A.  ±  180°, 
W—  the  dist.  of  centres  at  this  time, 
j  —  the  given  phase. 


<$,  180.]  eclipses.  403 

Lunar  eclipse. 
Find  W,  as  in  the  case  of  the  solar  eclipse,  by  Che  equations 

tang.  A  =z  -5-^—=  (869) 

b  R  cos.  d  v        " 

W=  D  cosec.  A.  (870) 

In  the  same  way,  find  another  value  of  W  for  another  time 
a  little  different  from  that  of  full  moon,  and  finish  the  compu- 
tation as  in  the  case  of  the  solar  eclipse. 

179.  Corollary.  The  same  method  might  be  used  if  longi- 
tudes and  latitudes  were  substituted  for  right  ascensions  and 
declinations. 

178.  Corollary.  This  eclipse  might  also  be  calculated  by 
the  process  of  §  169,  and  the  result  is  the  same  as  the  calcu- 
lation in  [B.  p.  417]. 

179.  Corollary.  At  the  beginning  or  end  of  the  eclipse,  we 
have 

J  =  U(r+*-o)  +  s-  (8H) 

180.  Problem.  Given  the  latitude  of  the  place  and 
the  apparent  time  of  the  beginning  or  end  of  a  phase  of 
a  solar  eclipse,  to  find  the  longitude  of  the  place. 

Solution.  From  the  supposed  longitude  of  the  place,  find 
the  Greenwich  time  of  the  observation,  and  for'this  time  find 
the  places  of  the  sun  and  moon,  and  their  relative  parallaxes  ; 
and  hence  the  Greenwich  time  of  the  phase.  The  difference 
between  the  Greenwich  time  and  the  observed  time  is  the 
longitude. 


404  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Longitude  from  solar  eclipse. 

181.  Corollary.  Instead  of  calculating  the  parallaxes  for  the 
rapidly  varying  positions  of  the  moon,  the  moon  ma\  be  supposed 
to  be  at  the  distance  of  the  stars,  while  the  sun  is  supposed  to 
be  at  a  distance  equal  to  the  real  distance  of  t  ie  moon.  But 
in  this  case  the  effect  of  parall  ix,  by  bringing  the  sun's  limb 
into  contact  with  the  moon's,  ca  \  only  be  obtained  by  supposing 
the  observer  to  be  in  the  situation  of  his  antipodes,  so  that  the 
parallax  may  be  reversed.  In  this  case  the  s  m's  diameter  must 
be  diminished,  just  as  the  moon's  diameter  is  rea  ly  increased. 

182.  Corollary.  The  sun  changes  its  place  so  slowly,  that 
its  right  ascension,  corrected  for  relative  parallax,  as  in  the 
preceding  corollary,  cannot  diifer  much  from  that  of  the  moon, 
at  the  time  of  conjunction  in  ri^ht  ascension.  For  a  time, 
then,  when  the  moon's  true  right  ascens  on  i>  nearly  that  of 
the  sun's  corrected  for  relative  parallax,  find  the  values  of 

R  s=  the  relative  right  ascension  -f-  parallax  in  R.  A. 

D  z=z  the  relative  declination  -f-  correction  for  declin. 

d0  —  the  sun's  declination  corrected  for  relative  paraL 
Rx  =  the  hourly  variation  of  R, 
D  ±  =  the  hourly  variation  of  2>, 

DQ  —  the  diff.  between  the  values  of  D  at  the  time  ty  and 
of  the  time  of  app.  conjunction  in  right  ascension, 
t0  z±  the  time  of  apparent  conjunction. 

Then  we  have,  by  (fig.  62.),  if  £  is  the  sun's  place  corrected 
for  relative  parallax,  and  BMA  the  moon's  true  orbit. 

*0  =  t  +  -J-,  D0  =  D  +  -|-  D,  =  SI,       (872) 
tan.  CMI=  tang.  *  =  jr^j  (873) 


<§>  184.]                             eclipses.  405 

Longitude  from  solar  eclipse. 

p  =  PS  =  DQ  cos.  i  (874) 

^~  m                         V         D n  cos,  i  #«**■% 

cos.  FSA  —  cos.  a  —  -^  z=  — -5 (875) 

^/  sin.  ^L#2        ^  sin.  (a  -f*  i)  /q-7«\ 

^.i  =  — : =z  ; ,  lc'D; 

sin.  ^17>S  cos.  i 


interval  of  moon's  passing  from  A  to  I 

AI  cos.  i        ^4  sin.  (a  -f-  *) 
~~  R  %  cos.  d  ~~        R  x  cos.  rf 


(877) 


time  of  obs.  phase  =  *0  +  S1"'  ^  +  *\  (878) 

J.\/ «    COS.  CL 

which  agrees  with  [B.  p.  463,  from  No.  12  to  the  end  of  the 
rule]. 

183.  Corollary.  The  diminution  of  the  semidiameter  re- 
quired by  §  181  is,  by  (753), 

da  =  o.P.  sin.  l"zin*JLs  (879) 

or  if  a  and  P  are  expressed  in  minutes, 

*<?  =  3600.  <r.  P.  sin.  1"  sin.  ^1 

—  T^.  tf  .p.  360000  sin.  1"  sin.  A.       (880) 

But  360000  sin.  I"  —  1.76,  (881) 

whence         1.76  sin.  A  is  the  factor  of  the  table  [B.  p.  443]. 

184.  Corollary.  The  latitude  of  the  moon  changes  so  slow- 
ly, that  its  latitude  at  the  time  of  the  phase  may  be  regarded 
as  the  same  with  its  latitude  at  the  supposed  time.  The  sun 
is  also  in  the  ecliptic  ;  so  that  if  we  put 

L  =  the  moon's  apparent  latitude, 

-^  —  the  apparent  relative  longitude, 


406  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Longitude  from  solar  eclipse. 

we  have 

J  =*/(J*  —L*)  =  */[{J+  L)(J  —  L)],  (882) 

from  which  the  apparent  and  true  longitude  of  the  moon  at 
the  time  of  conjunction  may  be  computed,  as  in  [B.  p.  413], 
and  thence  the  longitude  of  the  place. 

185.  Corollary,    If  both   the   beginning    and    end   of  the 
eclipse  are  observed,  and  if 

R  —  the  app.  relative  right  ascen.  at  the  beginning, 

R'  -=  that  at  the  end, 

D  zz  the  apparent  relative  dec.  at  the  beginning, 

D'  =z  that  at  the  end, 

we  have 

D  —  D 

The  dist.  gone  by  the  moon  =  E  =  (R1  —  R)  cos.  d  sec.  i. 

Then,  if  ^1  (fig.  62.)  is  the  place  of  the  moon  at  the  beginning, 
B  at  the  end  of  the  eclipse,  the  three  sides  AS,  AB,  and  BS 
of  the  triangle  ASB  are  given,  and  hence  AF,  ASF  can  be 
found,  whence 

a  ==  ASI=  ASF+  FSI, 

and  R  =  AI.  cos.  i  sec.  d  =z  — : —f^r  cos.  ^  sec.  a 

sin.  A  IS 

z=  J  sin.  «  sec.  d, 

whence  the  apparent  right  ascension  of  the  moon  at  the  in- 
stant of  the  commencement  of  the  eclipse  can  be  found,  and 
thence  its  true  right  ascension  and  the  Greenwich  time  of  the 
beginning. 


^  188.]  eclipses.  407 

Longitude  from  solar  eclipse. 

186.  Corollary.  The  method  of  the  preceding  corollary  is 
the  same  as  that  [B.  p.  407] ;  except  that  latitudes  and  longi- 
tudes are  used,  and  the  sun  is  in  the  ecliptic. 

187.  Corollary.  The  calculation  of  the  longitude  by 
means  of  occultations  is  the  same  as  that  by  means  of 
solar  eclipses,  as  in  [B.  pp.  410,  414,  446]. 


188.    Examples. 

1.    To  find   when  and  where  the  different  phases  of  the 
eclipse  of  Sept.  18,  1838,  begin  and  end  upon  the  earth. 

Solution. 

Greenw.  mean  t.  of  »ew  moon  =sa  7*56m38*.2 
J>  's  true  declination  —  D  =  2°  43'  52".3 
O's  true  declination        =  d     —  1°  49'  15".5 


D  —  d    =      54'36".8z=     3276".8 
3>'s  hourly  motion  in  dec.  z=  — 14'  10"  .5 

O's  hourly  motion  in  dec.  — —    0'  58". 3 


relative  motion  in  dec  =  Dx     ~  —  13'  12".2  ==  —  792".2 
J)  's  motion  in  right  ascension    ±z      26'    0".5 
O's  motion  in  right  ascension   z=        2'  14".  7 


rel.  motion  in  right  asc.  =  Rx  —      23'  45".8  =e     1425".8 


408  ♦  SPHERICAL    ASTRONOMY.  [cH.  XII. 

Solar  eclipse  of  Sept.  18,  1838. 

Dx  2.S98S3„  ar.co   7.101 17n 

R1  ar.  co.  6.84594 

D     sec.    0.00049  D—d  3.51545  3.55630 


-29°  5' 4"  tan.    9.74526n    cos.    9.94147   sin.  9.68673„ 


p  ==  2863//.7  3.45692  3.45692 

c  3.80112  3.80112 


*  — —  3518*  3.54638n 

_  __  58^  38s.   time  of  middle  =  7h  56m  38*  -—  t  =z  8h  55?n  1 6*. 
Now  for  the  phases,  we  have 
J)  's  equatorial  horizon,  paral.  =z 
Q's  equatorial  horizon,  paral. 

Relative  parallax  for  equator    = 
Reduction  for  lat.  45° 


Relat.  par.  for  lat.  45° 
J)'s  true  semidiameter 
Q's  true  semidiameter 

For  first  contact  J'  =  P>  +  5  +  o  =  84'  18".2=5058".2 

p.  3.45692 

*    ar.co.     6.29600  c     3.80112 


= 

53'  53" 

.7 

= 

8" 

.5 

=2 

53' 

45" 

.2 

== 

P' 

5" 

.3 

: 3219" 

ZZZ 

=  53' 

39" 

.9z= 

.9 

z=z 

s 

=  14' 

41/ 

'.2 

== 

a 

z=157 

57* 

.1 

k  =  55°  31'        cos.        9.75292  tan.     0.16314 


—  i  —  29°    5'  t  —  9210*  z=  2A  33™  20*  3.96426 


«  —  —  26°  26',  m.  t.  of  begin.  =  time  of  mid.  —  t  —  6*21*56' 
app.  t.  =  6h  21w  56s  +  5m  55*  -=  6h  27m  51*  =  96°  58' 

b  =  84°  36',  time  of  end  =  time  of  mid.  -f  t  =  1 P  2SCT  36' 
app.  t.  =  11*  28"  36*  +  6W  —  11*  34*  36*  =  173° 39' 


$  188.]  eclipses.  409 

Solar  eclipse  of  Sept.  18,  1838. 

a  cos.  9.95204  tan.    9.69647n 

d  cos.  9.99978  cosec.    1.49793 


lat.  =  63°3FN.  sin.  9.95182    266°  20'      tan.    1.19440 

at  beginning       long.  335  266°  20'— 96°  58'=  169°  22;  E. 

lat.  =a  63°  31'  +  9'  =  63°  40'  N. 
b  cos.  8.97363  tan.     1.02444 

d  cos.  9.99978  cosec.     1.49793 


lat  —  5°  24  N.    sin.  8.97341       90°  IF      tan.    2.52237, 

at  end  long.  =  173°  39'— 90°  1  F=83°  28'  W. 

lat.  =  5°  24/  +  2'  =  5°  26'  N. 
For  central  eclipse, 

P  3.45692 

P'      ar.co.     6.49215  c         3,80112 


A:  — 27°  13'     cos.      9.94907  tan.    9.71107 


— i  =  29°    5'        T  =  3252*  =  54™  12s      3.51219 


a  ==    1°  52'     time  of  begin.  ==  t.  of  mid.  -r-81  lm  4* 

b  =  56°  18' 

app.  time  =  8*  lm  4s  -J-  5m  57*  =  Sh  7m  Is  =  121°  45' 

time  of  end  =  time  of  middle  —  T  z=  9*  49771 28* 

app.  time  =  9"49w288+5w59'zz:9*55m27*:=  148°  52' 
a  cos.  9.99977  tan.  8.51310 

d  cos.  9.99978  cosec.  1.49793 


lat.  =  87°24/N.  sin.  9.99955  134°  17'     tan.  0.01103 

at  beginning      long.  =  134°  17'— 121° 45'=  12° 32' E. 
35 


410  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Solar  eclipse  of  Sept.  18,  1838. 

b.  cos.  9.74417  tan.  0.17593 

d.  cos.  9.99978  cosec.  1.49793 


lat.  =  33°  41'  N.  9.74395  91°  13'      tan.  1.67386 

at  end  long.  =  148°  52'  — 91°  13'  —  57°39/W. 

lat.  =  33°  41'  +  10'  =  33°  51'  N. 

2.  Find  the  places  where  the  eclipse  of  Sept.  18,  1838,  be- 
gins or  ends  in  the  horizon  af  8*  Greenwich  mean  time. 

Solution.  g  =  I  (P1  +  J)  =  4139" 

q>=  £(p/_^)  =919".l 

time  from  middle  eclipse  =  55m  16*  =  3316'  =  t 

t  3.52061 

c       ar.co.  6.19888  p.     3.45692 


k  =  27°  40'  tan.  9.71949  sec.  0  05273 


— t  =  29°  5'    J1  =  3233"    3.50965  ar.  co.  6.49035 


S=    1°15'    iJ'  —  9'==697".4         2.84348 

q  —  J  0  =  2522".5         3.40183 
P>  ar.  co.  6.49215 


2)19.22781 

■==     48°  32'  —  —  24°  16'  sin.     9.61390 


S-m=— 47°17/      cos.  9.83147  tan.  0.03465„ 

d  cos.  9.99978  cosec.  1.49793 


lat.  —  42°41/N.  sin.  9.83125   268°  19'    tan.  1.53158 


$   188.]  ECLIPSES.  411 

Solar  eclipse  of  Sept.  18,  1838. 

Lat.  =  42°  41'  +  11 '  =  42°  52'  N. 
Greenwich  app.  time  =:  8h  5m  57*  =:  121°  29' 
long.  =  268°  19'  —  121°  29'  =  146°  50'  E. 
and  at  this  place  the  eclipse  is  rising. 

S+  m  =  49°  47'         cos.  9.81002         tan.     0.07286 
d  cos.  9.99978       cosec.  1.49793 


Lat.  ==  40°  12'  N.  9.80980    91°  32'   1.57078„ 

long.  =  121°  29'  —  91°  32'  =  29°  57'  W. 
lat.  z=z    40°  12'  +        11'  =  40°  23'  N. 
and  at  this  place  the  eclipse  is  setting. 

3.  Find  the  place  on  the  southern  limit  of  the  eclipse  of 
Sept.  18, 1831,  which  corresponds  to  the  Greenwich  mean  time 
of  8*. 

Solution.  Since  the  altitude  is  not  known,  the  increase  of 
the  moon's  semidiameter  is  not  known,  but  it  may  be  supposed 
at  first  to  be  67/,  which  is  about  its  mean  value.     Hence 

j=o  +  s  +  6"  =.  30' 43 '.3  =  1844//.3 

p  —  ^  =  1019  "A 


j                      3.26583 
i              cos.     994147 

sin. 
sec. 

3.26583 
9.68673' 

u  =z  26'  52"            3.20730 

d+u  =  2°  16' 7"  =  D' 

0.00034 

R'  =  897".2  =  14'  57" 

2.95290, 

412  SPHERICAL    ASTRONOMY.  [CH.  XII. 


Solar  eclipse  of  Sept.  18,  1838. 

P 
c 
p  —  » 

ar.  co. 
ar.  co. 

49'  11" 

3.45692 

6.19888   P'.  ar.co. 

6.99166 

6.49215 
3.00834 

E 

t 

6.64746     W    cos. 
3.52061 

9.50049 

k'  =  55° 

0.16807             sec. 

0.25042 

—  i  =    29°   5'  4"     %  =?  34°  18'         sin.  9.75091 


M  =— 26°  44'  7"      cos.  9.95089      tan.  9.70219n 
Z=  tan.  9.83388 


6  =  31°  21' 2"         tan.  9.78477       sin.  9.71623 
a  +  D  —  33°  37'  9"         tan.  9.82274      sec.  0.07947 


h—R=  cos.  9.97950       tan.  9.49789n 


L  =z  32°  23'  tan.  9.80224 

For  a  more  accurate  determination. 

P1  =z  53'  45".2  —  3".2  =  53'  42"  =  3222" 
s  =  14'  41".2  +  11".6  =  14'  52".8 
j  =  s  +  a  =  30'  49".9  =  1849'.9 
15°  sin.  1"  ar.  co.  0.58204  0.58204 

Dx  2.89883n        Rx       3.15406 

B  =  —  3026      3.48087n       Z>  cos.  9.99951 


4  =       5440.1  3.73561 

For  8*  Gr.  time  we  have     D  =z  J>  's  dec.  =  2°  43'   5" 
i  R .  48"    2  log.     3.362     cZ  =z  Q's  dec.  =  1°  49'  12" 
1"  sin.     4.686     R  =  rel.  R.  A.  =  50".l 

2Z>  =  5°26'      sin.     8.976  D0  =  D 


2>0  —  2>  =  0  6.924     a;  =  2>0  —  rf  ==  53'  83"=3233' 


§  188.] 


ECLIPSES. 


413 


Solar  eclipse  of  Sept.  18,  1838. 


R. 

1.69984 

D 

COS. 

9.99951 

xo 

Vo  =  60" 

1.69935 

3.50961 

d 

sin. 

8.50187 

102".7 

8.50187 

2" 

0.20122 

2.01148 

B—  3026" 

A  = 

5446".  1 

0.00022 

■3028"  3.481 16„ 

5542".8ar.  co.  6.25628 


tan.  9.73744„ 
2  v  =  58°  40'   sec.  0.28396 


5542".8      3.74372 
P'      ar.  co.     6.49187 

sec.     0.05671 


•'==— 46°25/  tan.  0.02137„ 

2  (p  =  sec. 

<p  =  29°  20' 

%'  cos.     9.83848 

J  3.26926 


z 
2 


x    0.29252 

cos.     9.91703 
ar.  co.     9.69897 


tt=21'21"    3.10774 
rf  +  M  =  2°10,33" 


2)19.38019 

sin.     9.69004 
sin.     9.85996„ 
3.26926 

sec.     0.00032 


D  j±  2°43/  5"  —  R '=— 1347".5  3.12954 


x  =     32' 32"  =  1952", R—R=z  1297" '.4  3.11307n 
1>  cos.    9.99951 


3.11258?i 


35* 


414 


SPHERICAL    ASTRONOMY. 


[CH.  XII. 


Solar  eclipse  of  Sept.  18,  1838. 

y 

P>  ar. 

3.11258, 
co.          6.49187    x 

cosec.     0.25566 

3.11258, 
ar.  co.    6.71175 

COS. 

tan. 

tan. 
tan. 

cos. 

tan. 

M 

tan.    9.82433 

sin.     9.81834 
sec.     0.13824 

9.92002 

z 

sin.     9.86011 

0.02160 

s+D'=z 

41°  16'  24" 
43°  20' 

—  30°    8' 

—  31°  30' 8"        L. 

122°              long,  : 

9.94162 
9.97472 

k—R'  — 

tan.     9.78091 

zz    38°56'N. 
=  153°  20'  W. 

9.93250 

H  — 

9.90722 

lat.  ==    38°  56'  +  1 1'  =  39°  7'  N. 


4.  Find  where  the  solar  eclipse  of  Sept.  18,  1838,  is  central 
for  the  Greenwich  mean  time  of  9\ 


t  =  4W  44s  = 
P 

284s 

3.45692 

sec.  0.00044 

c  ar.  co. 
tan. 

tan. 
cos. 

tan. 
tan. 

cos. 

tan. 

2.45332 
6.19888 

Jc=z2°  34' 

8.65220 

J' 

P                  ar 

3.45736 
.  co.      6.49215 

Z 

s  z=  31°  39' 

sin.  9.94951 
tan.  9.78987 

0.29106 
9.93007 

4:=  59° 

d  +  d  —  60°  49' 

sin.  9.93304 
sec.  0.31193 

0.22113 
0.25298 

h  =z  47°  18' 

tan.  0.03484 
L  =z  50°  32' 

9.83140 

0.08438 

§  188.]  eclipses.  415 


Solar  eclipse  of  Sept.  18,  1838. 


For  a  more  accurate  determination. 
P>  —  53'  45".2  —  6".2  =  53'  39 "  ==  32197 
D  =  2°  28'  54",        df  =  1°  48'  13",        D  cos.  9.99959 
R  —  25'  5".4  =  1505  ".4  3.17765 

x0  =  2>  —  d=40'41"  —  2441"      3.38757  6.61243 


£  cos.  9.93013     sec.  0.06987  tan.  9.78967 


j'  3.45744 

P'.  ar.  co.  6.49228 


tan.  0.29208     sin.  9.94972 


q  =z59°  3' 24"  tan.  0.22221  sin.  9.93332 

4  -f  <Z  =  60°  51'  37'  tan.  0.25375  sec.  0.31252 


:47°20/  cos.  9.83100  tan.  0.03551 


Z,z=50°33'  tan.  0.08475 

lat.  bz  50°  33'  +11'  =  50°  44'  N. 
long.  =  47°  20'  —  136°  29'  =  89°  9'  W. 

Calculate  the  solar  eclipse  of  September  18,  1838,  for  the 
city  of  New  York. 

Calculation  for  9^  Gr.  mean  time,  by  the  principles  of  §  182. 
Gr.  app. t.  —9h5m57sA    iz=:40o 4240"— 1120"— 40° 31'20" 
N.Y.  long.  =4ft 56m  4S.5 

h        =z4A9wl52*.9P.M.  P'  =  53'  45".2— 4".6  =  53'  40"6. 


416  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Solar  eclipse  of  Sept.  18,  1838. 

P>.  =  3220".6  3.50794 

L.  cos.  9.88090  9.88090         9.88090 

15  ar.  co.  8.82391 

2  h  =  Sh  19"*  45*.8    sin.  9.94782 


S'  2.16057 

D  sec.  0.00041 


2"  24'.  9  2.16098 

2  h'=8h  17m21%  A=4A8wl40*.5,  log.Ris.  4.72684  sin.  9.94661 
£  — d=38°43'7"N.   cos.  78022  d  cos.  9.99979 


40507  4.60753 


22°  2'  N.  sin.  37515     cos.  9.96706  sec.  0.03294 


M  cos.  9.83794  sin.  9.86045 

P'.  3.50794 


dd  =  34'  15".6  N.  3.31294  £'   2.16057 

9  =  2°  22'  29"  cos.   9.99963  sec.  0.00037 


sR=z2m  24*.9  =  36'  13"  =  21 73"  2.16094 

JR— *JK  ;=  —  668"  2.82478„ 

D  —  d<  =  &  26"  =  385"  ar.  co.     7.41454         2.58546 


8  =  —  60°  1'  tan.    0.23895„  sec.  0.30125 

W  2.88671 


■§>  188.] 


ECLIPSES. 


417 


Solar  eclipse  of  Sept.  18, 

1838. 

Dx  =  792".7 

2.8991  ln 

RL  z=z  1425" 

ar.  co.    6.84619 

d 

sec.    0.00037 

i  =  —  29°  6' 

tan.    9.74567n  W.  2.88671 

iST-f-  f  =  —89°  7' 

cos.    8.18798 

JP- 

1.07469 

For  beginning  of 

eclipse. 

^  =  30'  38".3  : 

=  1838".3 

ar.  co.  6.73565 

P 

sec.  2.18966 

1.07469 

k  =  89°  38' 

cos.  7.81034 

W. 

2.88691 

3600' 

3.55630 

l  ZZl 

cos.  9.94137 

^ 

ar.  co.  6.84619 

tf 

sec.  0.00037 

a=  31' 

sin.  7.95508 

t,  —  —  40OT 

3.37568 

mean  time  at  N.  Y.  3ft  24771, 

For  a  more  accurate  calculation. 

Gr.  mean  time  —  8h  207* 
15  P'  cos.  L  2.21275 

2h=  6"  59m44*.6  sin.  9.89928 


Gr.  m.  t.  s=  8*  20w. 

h  =  3*  29™  52'.3 
d  —  1°  48'  6SKUI 
R  =  555".0 


£'  2.11203 

2>—  2°38'21".l   sec.  0.00046 


2m  98.6 


2.11249 


418  SPHERICAL    ASTRONOMY.  [cH.  XII. 

Solar  eclipse  of  Sept.  18,  1838. 

L         cos.  9  88090    9.88090 

2A'  =  6*57OT35*/i'  =  3/i28m47*.5  log.  Ris.  4.58778  sin.  9.89770 
£.— d=38042'27"  N.  cos.  .78033  d.  cos.  9.99978 


.29407      4.46846 


29°  5'  42"     N.  sin.  .48626   cos.  9.94142  sec.  0.05858 


M  cos.  9.86114  sin.  9.83718 

P>.   3.50794 


&d=.  34'  4".l  N.  3.31050  S'   2.11203 

d'  =  2°  22'  57"  cos.  9.99962  sec.  0.00038 


d  R  =  2m  9s  55  =  32'  23".2  =  1943".2         2. 1 1243 
R  —  6  R  =  —  1388".2         3.14245„ 
D—d  =  15'  24".l=924'.l  ar.  co.  7.03428    2.96572 


S  =  —56°  19'  32''     tan.  0.17635„  sec.  0.25612 


Dx=  —  792.5      2.89900„  W.  3.22184 
Rx  =  1425.3  ar.  co.  6.84610 
d'.  sec.  0.00038 


i  =  —  29°  5'  48"    tan.  9.74548n 
8  -f  i  =  —  85°  25'  20"  cos.  8.90207 


2.12391 
4  =  30'  30".8  =  1830".8  ar.  co.  6.73736 


&  =  85°  50'  cos.  8.86127 


§  188.]  eclipses.  419 

Solar  eclipse  of  Sept.  18,  1838. 


k 

sec. 

1.13873 

w 

3.22184 

3600" 

3.55630 

i 

cos. 

9.94141 

Rx   cos.  d 

ar.  co. 

6.84648 

a=     24' 

40" 

sin. 

7.85583 

1 1  =  —  12*  28*  2.56059 

mean  time  at  N.  Y.  3*  17m  20%     Gr.  time  =  8h  13w  24*. 

For  a  still  more  accurate  calculation. 
Gr.  mean  time         Sh  13m  30*  h  =  3*  23m  22*.3 

15  P<  cos.  L  2.21279  d  =  1°  48'  59".3 

2  ft=6M6m44*.6  sin.  9.88954  R  s=  401.1 


8'  2.10233 

D  =  2°  39'  53".2  sec.  0.00046 


2W6*.7  2.10279   L.  cos.  9.88090  9.88090 

2h'  =  &Mm37\9  h'=3h22m18>  log.  Ris.  4.56222  sin.  9.88790 
£,— d=38°  42'  21"  N.  cos.  78035  d  cos.  9.99978 


27727  4.44290 


30°  12'  7"  N.    sin.  50308      cos.  9.93664  sec.  0.06336 


M.  cos.  9.86553  sin.  9.83216 

P'  3.50798 


dD=:     34'2".4N.  3.31015 


420  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Solar  eclipse  of  Sept.  18,  1838. 

Sf  2.10233 
d'  ss  2°  23'  1".7  cos.  9.99962  sec.  0.00038 


S  R  =  2*  6ff.68  =  31'  40".2  =  1900 '.2  2.10271 

R  —  9  R  =  —  1499".l  3.17593n 

D  —  tf '  =  16'  51  ".5  =  1011  ".5  ar.  co.  6.99503         3.00497 


S=z  —  55°  58'  2"  tan.  0.17048  sec.  0.25207 


W.   3.25704  3.25704 

jSf+t  =  — 85°3'50"  cos.  8.93472 

(3600 "cos.*)  -±  Rx  cos.  d  =  0.34419^  ar.co.  6.73753 


*  =  85°7'32"  sec.    1.07071        cos.  8.92929 

a=         3  42"  sin.    7.03193 


t^—  50s. 6  1.70387 

Mean  Gr.  time  ==  Sh  12m  39*4.,     N.  Y.  m.  t.  =  3h  16m34*.9. 

For  beginning  of  annular  phase. 

J  =  V  15'.9  =:  75".9  ar.  co.  8.11976 

W.  2.88671  p         1.07469 

k  =  Sr  sec.  0.80555  cos.  9.19445 

W  3600"  cos.  z  3.49767 

JR1  cos.  d        ar.co.  6.84656 

a=  —  8°  7'  sin.  9.14980n 


ft  =  25™  36s  3.18629n 

Gr.  mean  time  ac  9*  25m  36*. 


§  188.]  eclipses.  .    421 

Solar  eclipse  of  Sept.  18,  1838. 

For  a  more  accurate  calculation. 

Gr.  mean  time  —  9^  30OT,  h  =  4h  39™  53s.4 

15  P  cos.  L  2.21266  d  =  1°  47'  44".7 

2A  =  9?l19™46s.8   sin.  9.97291  jR  =  2217".9 


£'  2.18557 

2>  =  2021'48".2  sec.  0.00037 


2™33s.4  2.18594   L.  cos.  9.88090         9.88090 

2A/z=9A17?ra138.4  A/=4*38w36*.71og.Ris.  4.81444  sin.  9.97202 
L— d=38°  43'  35"  N.cos.  78015  d  cos.  9.99979 


49560 

cos. 

COS. 

pi 

4.69513 

16°  31'  56" 

N.  sin.  28455 

9.98166  sec.  0.01834 

M 

9.82528  sin.  9.8712C 
3.50785 

3d  =  34'  24 "A  3.31479  S'.  2.18557 

61  =  2°  22'  19".l  cos.  9.99963  sec.  0.00037 


>  JR  =5  2301"  ==  13s.4  a  #  =  2.18594 

R  —  *R=z—  83".  1  1.91960n 

Z>  —   d'zn— 30".9  ar.co.  8.45967n       l.48996n 


-ST  —  — - 112°  40'  54"        tan.  0.37890  sec.  0.41 385„ 


D  ,  =  —  793"  2.89927n  W  1.90381 

Rx  =  1423".2    ar.  co.  6.84674 
d'  sec.  0.00037 


i  ±2  —  29°  8'  50"  tan.  9.74638n 

36 


422  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Solar  eclipse  of  Sept.  18,  1838. 

W    1.90381 
S+i  =  —  141°  49'  44"  cos.  9.89551, 


W        1.90381  1.79932n 

3600"  -f-  Rx  cos.  d'  0.40341      ^    ar.co.    8.14997 


k  ss  152°  51'  9"       sec.  0.05071  cos.  9.94929n 

a  ss    17°  1'  25"        sin.  9.28151 
•  cos.  9.94120 


*!  ss  38M  1.58064 

Gr.  m.  time  =  9*  30m  38M.     N.  Y.  m.  t.  =  4*  34w  33».6. 

In  the  same  way  we  should  find  for  the  end  of  the  annular 
phase. 

N.  Y.  mean  time  ss  4*  38m  12% 

and  for  the  end  of  the  eclipse, 

N.  Y.  mean  time  ss  5*  47m  54% 

5.  If  the  beginning  of  the  solar  eclipse  of  Sept.  18,  1838,  had 
been  observed  at  New  Orleans,  in  1  at.  29°  57' 45"  N.,  at  2h 
19OT  P.5  mean  time,  what  would  be  the  longitude  of  New  Or- 
leans ? 

Solution.    Let  the  supposed  long,  zs  6* 

Greenwich  mean  time  =.  Sh  19w  P.5,  h  ss  2*  24OT  58' A 

L  ss  reduced  lat.  =  29°  57'  45"  —  9'  55"  ss  29°  47' 50" 

P'=  53'  45".3  —  2".8  ss  53'  427.5  =  3222".5 

d  ss  1°  48'  53".3 

R=z    530".l 


$  188.]  eclipses.  423 

Solar  eclipse  of  Sept.  18,  1838. 

P'  3.50820 

15.    ar.  co.  8.82391 

L.     cos,     9.93841     cos.  9.93841    cos.  9.93841 
2h=z4/l49m56s.8  sin.  9.77174 


8'  2.04226 

D  =  2°  38'  34".8  sec.  0.00046 


23h'—lm50\3  2.04272 

2h'=z4:h48m6s.5,7i<  —  2h24m3s.2l  R.  4.28132    sin.  9.76936 
L—d=27°  58' 57"  N.cos.  89115  d.  cos.  9.99978 


16577 

4.21951 

cos.  9.84045 

cos.  9.83005 
P'.   3.50820 

sec. 
sin. 

sec. 

sec, 
W 

cos. 

46°  10'  3"   N.  sin.  72, 

538 

ar 

'.  CO. 

sec. 
tan. 

0.15955 

M 

9.86732 

sd=z  25' 9" 
d'—2°  U'2'3 

3.17870 
cos.  9.99967 

3.05082n 
.  co.  6.83194 

2.04226 
0.00033 

iR  =  1654  ".5  ==  1110s.3 
Rz=  sR  —  —  1124,"  A 
D  —  d'   =  1472  ".5 

2.04259 
3.16806 

S  =  —  37°  20'  16" 

tan.  9.88243,, 

2.89894 
6.84607 
0.00033 

0.09960 

Dx  =•  — 792<».4 

22,  =  1425".4     ai 
d' 

3.26766 

i   =3  —  29°  5'  20" 

£  -f  i  =  —  66°  25'  36" 

9.74534 

960198 

2.86964 

424  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Solar  eclipse  of  Sept.  18,  1838. 

i  cos.  9.94145  p.  2.86964 

W.    3.26766  J.  ar.  co.  6.73824 


k  —  66°  tit  2"  sec.  0.39212  cos.  9.60788 

a  =  —20'  34"  sin.  7.77689 

Rx  cos.  d'  ar.  co.  6.84640 

3.55630 


t  =  60-.37  =  lm  0*.37         1.78082 
long.  =  6h  lm  0*.37  5=  90°  15'  #'. 

For  a  more  accurate  calculation. 
Gr.  mean  time  =  8h  20w 
df  =  1°  48'  52".8,     D  =  2°  38'  21".l 
JR  =  555' .4, 

tf '  =  2°  14'  1".8  cos.  9.99967 

R  —  dR=z  —  1099  ".1  3.04 104n 

D  —  d=      1459".3       ar.  co.  6.835S5  3.16415 


£'  z=  —  36°  57'  52"     tan.  9.87656n        sec.  0.09745 

3.26160  3.26160 

S  +  i  =  —  66°  3'  12"  cos.  9.60840 

3600  cos.  i  ~  J^  cos.  eZ'  0.34415  2.87000 

a  =z  42"       sin.  6.30882  ^.  ar.  co.  6.73821 


66°  3'  54"      sec.  0.39179         cos.  9.60821 


*x  =  —  2*.02  0.30636 

*ong.=      90°  14' 7"  W. 


§  188.]  eclipses.  425 

Solar  eclipse  of  Sept.  18,  1838. 

6.  To  find  the  times  of  the  beginning  and  end  of  the  annu- 
lar eclipse  of  Sept.  18,  1838,  at  Washington,  D.  C,  and  the 
times  of  the  formation  and  rupture  of  the  ring. 

Extracts  from  the  Nautical  Almanac. 

Greenwich  mean  time,  of  Sept.  1838. 

mO*  ©'s  R.  A.  =  11"  38™ 25*.08,  Dec.  ■=. 2°  20'  14'.3 N. 

18  .0   ©'s  R.  A.  =  11  42     0  .61,  Dec.  =  1  56  58  .2  N. 

19  .0   ©'s  R.  A.  =  11  45  36  .16,  Dec.  =  1  33  39  .5  N. 

20  .0   ©'s  R.  A.  =  ,11  49  11  .75,  Dec.  =  1  10  18  .6  N. 

17  .0  equation  of  time  sfs  5m28ff.55 

18  .0  equation  of  time  =  5   49  .57 

19  .0  equation  of  time  =  6    18  .57 

20  .0  equation  of  time  =  6   31  .53 

18*  6h  3)  'i  R.  A.  =  III  39w49t.64,  Dec.  =  3°  11'  24", 6  N. 
18  7  ])'sR.A.zzll  41  33  .74,  Dec.  =  2  57  14  .9  N. 
18  8  3)  'sR.  A.  =  11  43  17  .79,  Dec.  =  2  43  4  .6  N. 
18  9  2)  's  R.  A.  =  11  45  1  .79,  Dec.  =  2  28  53  .9  N. 
18  10  D's  R.  A.  =:  11  46  45  .75,  Dec.  =  2  14  42  .6N. 
18  11  D  's  R.  A.  =  11  48  29  .67,  Dec.  —  2  0  31  .0  N. 
18     0    D  's  semid.  =  14'  4 1  ".5  Hor.  Par.  =  53'  54".8 

18  12    3)  's  semid.  =  14'  41  ".1  Hor.  Par.  =  53'  53  .3 

©'s  semid.  —  15'  57".  1  Hor.  Par.  —  8".5 

Ans.     Beginning  of  eclipse  at  3*  5™15\6  W.  mean  time, 
of  ring      at  4  23  46  .1 
end  of  ring      at  4  29  42  .3 
of  eclipse  at  5  39  30  3 


426  SPHERICAL    ASTRONOMY.  [CH.  XII. 

Solar  eclipse  of  May  15,  1836. 

7.  Calculate  the  time  of  the  beginning  and  end  upon  the 
earth  of  the  solar  eclipse  of  May  15,  1836,  and  the  places 
where  it  is  first  and  last  seen. 

Extracts  from  the  Nautical  Almanac. 

Greenwich  mean  time  of  May,  1836. 

Ud  0h  ©'s  R.  A.=  3/l  25-  5U3  Dec.  =  18°  42'  21".0N. 


15 

0 

3  29  1  .93 

18  56  35  .9  N. 

16 

0 

3  32  59  .30 

19  10  31  .6N. 

17 

0 

3  36  57  .25 

19  24  7  .8  N. 

14  0     equation  of  time  z=  3m56s.30 

15  0     equation  of  time  =  3  56  .05 

16  0     equation  of  time  zz  3  55  .24 

17  0     equation  of  time  =  3  53  .86 

14  22  3)'sR.A,z=3  20  41  .89  Dec.  s=  18  40  52  .9  N. 
3  22  41  .69  18  51  11  .4  N. 
3  24  41  .69  19  1  24  .7  N. 
3  26  41  .88  19  11  33  .0  N. 
3  28  42  .26  19  21  36  .1  N. 
3  30  42.84  19  31  34  .ON. 
3  32  43  .62  19  41  26  .6  N. 
3  34  44  .60                19  51  13  .9  N. 

15  6  3  36  45  .77  20     0  55  .9  N. 
14*  12*  J)  's  semid.  =  14'  52'  .3     Hor.  Par.  =  54'  34' .5 
15     0     J)  's  semid.  =  14  49.9     Hor.  Par.  =  54  25  .6 
15  12    3>'s  semid.  =  14  47  .7     Hor.  Par.  =  54  17  .7 

<g)'s  semid.  —  15'  49'.9  Hor.  Par.  =  8".5 


14  23 

15 

0 

15 

1 

15 

2 

15 

3 

15 

4 

15 

5 

§  188.]  eclipses.  427 

Solar  eclipse  of  May  15,  1836. 

Ans.    Time  of  begin.  z=  Ud  23*  6-  30' 

in  Long.  z=  76°  53'  W.,  and  Lat.  =z  2°  10'  S. 
Time  of  central  begin.  =  15d  0*  18"  10* 

in  Long.  =  98°  16'  W.,  and  Lat.  =  7°  58'  N. 
Time  of  central  end  =z  15*  3*  44™  44* 

in  Long.  =  52°  41'  E.,  and  Lat.  =  44°  50'  N. 
Time  of  end  =  15d  4A  56m  24* 

in  Long.  =  28°  51'  E.,  and  Lat.  =  35°  13' N. 

8.  Find  where  the  solar  eclipse  of  May  15,  1836,  is  rising 
or  setting  at  15d  0h  S0m  Greenwich  mean  time. 

Ans.     In  Long.  =z    90°  44'  W.     Lat.  =-  21°  32'  S. 
and  in  Long.  =  116°  42'  W.     Lat.  =  42°  29' N. 

9.  Find  the  place  which  is  upon  the  southern  limit  of  the 
above  eclipse's  visibility  at  the  Greenwich  mean  time  of  15d  3\ 

Ans.     Long.  =  10°  11'  36"  W.     Lat.  =  21°  41'  24"  N. 

10.  Find  the  place  which  is  upon  the  northern  limit  of  the 
visibility  of  the  annular  phase  of  the  solar  eclipse  of  May  15, 
1836,  at  the  Greenwich  mean  time  of  3*  11™  27*. 

Ans.     Long.  ==  3°  8'  30"  W.     Lat.  s=  56°  30'  54"  N. 

11.  Find  the  place  where  the  solar  eclipse  of  May  15,  1836, 
is  central  at  3A  11™  27*  Greenwich  mean  time. 

Ans.     Long.  =  3°  53'  18"  W.     Lat.  =  58°  29'  18"  N. 

12.  Find  where  the  solar  eclipse  of  May  15,  1836,  is  cen- 
tral at  noon. 

Ans.     Long.  =  36°  20'  W.     Lat.  =s  49°  17'  N. 


428  SPHERICAL    ASTRONOMY.  [CH,  XII. 

Solar  eclipse  of  May  15,  1836. 

12.  Calculate  the  time  of  the  beginning  of  the  eclipse  of 
May  15,  1836,  for  the  Observatory  of  Edinburgh. 

Ans.     Beginning  of  eclipse  =  lh  32m  40s  Edinb.  M.  time. 

13.  Suppose  the  beginning  of  the  solar  eclipse  of  May  15, 
1836,  to  be  observed  to  take  place  at  1*  36™  35s.6  app.  time, 
in  latitude  55°  57'  20"  N.,  and  longitude  about  I2m  W. ;  find 
the  longitude  of  the  place  of  observation. 

Ans.     Long.  =  12-  43s.  7  W. 


THE    END. 


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